The stock price process $(X_t)$ is a geometric Brownian motion with drift $\mu=0$. Thus, $$X_t=X_0\exp\left(-\frac{1}{2}\sigma^2t+\sigma W_t\right).$$
Assume you have constant interest rates $r_t\equiv r$ and are interested in a European-style claim, then, using risk-neutral pricing, the time zero price of a claim paying $H=X_T^2$ equals
$$ V_0 = e^{-rT}\mathbb{E}^\mathbb{Q}[X_T^2].$$
The time $t$ price would simply read as $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[X_T^2\mid\mathcal{F}_t]$.
So, you need the moments of $(X_T^2)$ under the risk-neutral measure $\mathbb{Q}$. Regardless what mean $(X_t)$ has in the real-world, its drift in the risk-neutral world is $r$, (potentially $r-q$ where $q$ is the (constant) dividend yield). Thus, under $\mathbb{Q}$,
$$\ln(X_t) \overset{\mathrm{Law}}{=} \ln(X_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t.$$
As you see, $(X_t)$ is log-normally distributed. In general, if $\ln(Y)\sim N(m,s^2)$, then $\mathbb{E}[Y^k]=e^{km+\frac{1}{2}k^2s^2}$ for all $k\geq1$, see here. Thus, putting everything together, and using that $\ln(X_0)=0$, you obtain as price of your power contract
\begin{align*}
V_0 &= e^{-rT}\mathbb{E}^\mathbb{Q}[X_T^2] \\
&= \exp\left(-rT+2\left(r-\frac{1}{2}\sigma^2\right)T+2\sigma^2T\right) \\
&= \exp\left(\left(r+\sigma^2\right)T\right).
\end{align*}
As a matter of fact, if $\gamma>0$, you can show that $(X_t^\gamma)$ is again a geometric Brownian motion using Ito's Lemma. Furthermore, you get that the time $t$ price of a European-style claim paying $X_T^\gamma$ is given by
\begin{align*}
V_t = (X_t)^\gamma\cdot\exp\left((\gamma-1)\left(r+\frac{1}{2}\gamma\sigma^2\right)(T-t)\right).
\end{align*}
Indeed, setting $t=0$, $X_0=1$ and $\gamma=2$ recovers the above solution.