By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$
so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$
In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}x_0^2 $$