Step 1: Know your distribution

We have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( \left(r-\frac{1}{2}\left(\frac{1}{3}\sigma^2 t^2\right)\right)t + \sqrt{\frac{1}{3}\sigma^2t^2} W_t \right), \end{align*} as shown here. In particular, the stock price is log-normally distributed for every time point $t$.

Step 2: Remember your toolkit

We used the following result: If $\ln(X)\sim N(m,s^2)$, then \begin{align*} \mathbb{E}[\max\{X-K,0\}] &= e^{m+\frac{1}{2}s^2}\Phi\left(\frac{m-\ln(K)+s^2}{s}\right)-K\Phi\left(\frac{m-\ln(K)}{s}\right). \end{align*} In your example, \begin{align*} \ln(S_T) &= \ln(S_0) + rt-\frac{1}{6}\sigma^2 T^3 + \sqrt{\frac{1}{3}\sigma^2 T^3} Z, \\ \implies \mathbb{E}[\ln(S_T)] &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3, \\ \implies \mathbb{V}\mathrm{ar}[\ln(S_T)] &= \frac{1}{3}\sigma^2 T^3. \\ \end{align*}

Step 3: Put everything together

Assuming the absence of arbitrage, the option price is then the discounted expected payoff. I'll assume that the above stock price dynamics are with respect to the risk-neutral measure. Then,

\begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}] \\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT+\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT-\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)\\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r+\frac{1}{6}\sigma^2 T\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r-\frac{1}{6}\sigma^2 T^2\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right). \end{align*}

Step 1: Know your distribution

Since $\int_0^t W_s\mathrm{d}s\sim N\left(0,\frac{1}{3}t^3\right)$, we have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( \left(r-\frac{1}{2}\left(\frac{1}{3}\sigma^2 t^2\right)\right)t + \sqrt{\frac{1}{3}\sigma^2t^2} W_t \right), \end{align*} where $Z\sim N(0,1)$, as shown here. In particular, the stock price is log-normally distributed for every time point $t$.

Step 2: Remember your toolkit

We'll use the following result: if $\ln(X)\sim N(m,s^2)$, then \begin{align*} \mathbb{E}[\max\{X-K,0\}] &= e^{m+\frac{1}{2}s^2}\Phi\left(\frac{m-\ln(K)+s^2}{s}\right)-K\Phi\left(\frac{m-\ln(K)}{s}\right). \end{align*} In your example, \begin{align*} \ln(S_T) &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3 + \sqrt{\frac{1}{3}\sigma^2 T^3} Z, \\ \implies \mathbb{E}[\ln(S_T)] &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3, \\ \implies \mathbb{V}\mathrm{ar}[\ln(S_T)] &= \frac{1}{3}\sigma^2 T^3. \\ \end{align*}

Step 3: Put everything together

Assuming the absence of arbitrage, the option price is then the discounted expected payoff. I'll assume that the above stock price dynamics are with respect to the risk-neutral measure. Then,

\begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}] \\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT+\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT-\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)\\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r+\frac{1}{6}\sigma^2 T\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r-\frac{1}{6}\sigma^2 T^2\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right). \end{align*}

Step 1: Know your distribution

We have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( \left(r-\frac{1}{2}\left(\frac{1}{3}\sigma^2 t^2\right)\right)t + \sqrt{\frac{1}{3}\sigma^2t^2} W_t \right), \end{align*} as shown here. In particular, the stock price is log-normally distributed for every time point $t$.

Step 2: Remember your toolkit

We used the following result: If $\ln(X)\sim N(m,s^2)$, then \begin{align*} \mathbb{E}[\max\{X-K,0\}] &= e^{m+\frac{1}{2}s^2}\Phi\left(\frac{m-\ln(K)+s^2}{s}\right)-K\Phi\left(\frac{m-\ln(K)}{s}\right). \end{align*} In your example, \begin{align*} \ln(S_T) &= \ln(S_0) + rt-\frac{1}{6}\sigma^2 T^3 + \sigma \sqrt{\frac{1}{3}T^3} Z, \\ \implies \mathbb{E}[\ln(S_T)] &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3, \\ \implies \mathbb{V}\mathrm{ar}[\ln(S_T)] &= \frac{1}{3}\sigma^2 T^3. \\ \end{align*}

Step 3: Put everything together

Assuming the absence of arbitrage, the option price is then the discounted expected payoff. I'll assume that the above stock price dynamics are with respect to the risk-neutral measure. Then,

\begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}] \\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT+\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}}\sigma T\sqrt{T}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT-\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}}\sigma T\sqrt{T}}\right)\\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r+\frac{1}{6}\sigma^2 T\right)T}{\sqrt{\frac{1}{3}}\sigma T\sqrt{T}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r-\frac{1}{6}\sigma^2 T^2\right)T}{\sqrt{\frac{1}{3}}\sigma T\sqrt{T}}\right). \end{align*}

How do we know the above formula?

Step 1: Know your distribution

We have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( \left(r-\frac{1}{2}\left(\frac{1}{3}\sigma^2 t^2\right)\right)t + \sqrt{\frac{1}{3}\sigma^2t^2} W_t \right), \end{align*} as shown here. In particular, the stock price is log-normally distributed for every time point $t$.

Step 2: Remember your toolkit

We used the following result: If $\ln(X)\sim N(m,s^2)$, then \begin{align*} \mathbb{E}[\max\{X-K,0\}] &= e^{m+\frac{1}{2}s^2}\Phi\left(\frac{m-\ln(K)+s^2}{s}\right)-K\Phi\left(\frac{m-\ln(K)}{s}\right). \end{align*} In your example, \begin{align*} \ln(S_T) &= \ln(S_0) + rt-\frac{1}{6}\sigma^2 T^3 + \sqrt{\frac{1}{3}\sigma^2 T^3} Z, \\ \implies \mathbb{E}[\ln(S_T)] &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3, \\ \implies \mathbb{V}\mathrm{ar}[\ln(S_T)] &= \frac{1}{3}\sigma^2 T^3. \\ \end{align*}

Step 3: Put everything together

Assuming the absence of arbitrage, the option price is then the discounted expected payoff. I'll assume that the above stock price dynamics are with respect to the risk-neutral measure. Then,

\begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}] \\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT+\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT-\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)\\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r+\frac{1}{6}\sigma^2 T\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r-\frac{1}{6}\sigma^2 T^2\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right). \end{align*}