The integral $I_T$ is an Itô stochastic integral therefore its expectation is $0$. This is because $I_T$ is a martingale (see e.g. Theorem 4.3.1 in Shreve), hence:
$$\mathbb{E}[I_T]=I_0=0$$
You can also see this by considering the definition of a stochastic integral, which involves the sum of terms of the form $f(W_{t_i})(W_{t_{i+1}}-W_{t_i})$, and using the independence of Brownian increments $W_{t_i}-W_0$ and $W_{t_{i+1}}-W_{t_i}$.
From the above, we get:
$$\mathbb{V}[I_T]=\mathbb{E}[I_T^2]$$
Given $I_T$ is an Itô integral and that the process $Z_t\triangleq \sqrt{|W_t|}$ is adapted to the filtration generated by $W_t$, by Itô's Isometry:
$$\begin{align}
\mathbb{E}[I_T^2]&=\mathbb{E}\left[\int_0^TZ_t^2\text{d}t\right]
\\[3pt]
&=\int_0^T\mathbb{E}[|W_t|]\text{d}t
\end{align}$$
$W_t$ is normally distributed. By symmetry of the Normal distribution, the expectation of $|W_t|$ is equal to twice the expectation of $1_{\{W_t\geq0\}}W_t$, namely:
$$\begin{align}
\mathbb{E}[1_{\{W_t\geq0\}}W_t]&=\int_0^\infty w\frac{1}{\sqrt{2\pi t}}e^{-\frac{w^2}{2t}}\text{d}w
\\[3pt]
&=\int_0^\infty v\frac{1}{\sqrt{2\pi}}e^{-\frac{v^2}{2}}\sqrt{t}\text{d}v
\\[3pt]
&=\sqrt{\frac{t}{2\pi}}\int_0^\infty ve^{-\frac{v^2}{2}}\text{d}v
\\[8pt]
&=\sqrt{\frac{t}{2\pi}}
\end{align}$$
where we've made the change of variables $v=w/\sqrt{t}$. Thus $\mathbb{E}[|W_t|]=\sqrt{2t/\pi}$, from which it comes:
$$\begin{align}
\mathbb{V}[I_T]&=\sqrt{\frac{2}{\pi}}\int_0^T\sqrt{t}\text{d}t
\\[6pt]
&=\sqrt{\frac{8}{9\pi}}T^{3/2}
\end{align}$$
References
Shreve, S. (2004). Stochastic Calculus for Finance II, Springer.
