$W_{t}$ is a wiener process and $a^{+} = max (a, 0)$ So far I've had one thought. Use Ito's formula somehow. Here's what I got: For $F(t, X(t)) = x^{2}$ $$ \left(\int_{0}^{t}W^{+}_{s}dW_{s}\right)^{2} = \int_{0}^{t}\left(W^{+}\right)^{2}_{s}dt +2 * \int_{0}^{t}\left(\int_{0}^{t}W^{+}_{s}dW_{s}\right)W^{+}_{s}dW_{s} $$
Which is not much better, but somewhat of an equation. Though I might be in the wrong woods here, trying to find the integral, when I should be focusing on finding the variance directly
By defintion $\mathbb{E} (\int_0^t W_s^+dW_s)=0$, so we only need to compute the second moment. This follows easily from usage of the Ito Isometry formula: We have that: $$\mathbb{E} (\int_0^t W_s^+dW_s)^2=\mathbb{E}\int_0^t (W_s^+)^2ds=\int_0^t \mathbb{E}(W_s^+)^2ds$$ We now note that we have that $W_s\sim s^{1/2}W_1$ so that: $$\int_0^t\mathbb{E} (W_s^+)^2sds=\mathbb{E} (W_1^+)^2\int_0^tsds=\mathbb{E} (W_1^+)^2\frac{1}{2}t^2$$
To compute the remaining term, we have note that as $W_1\sim \cal{N}(0,1)$, we have that: $$\mathbb{E} (W_1^+)^2=\frac{1}{2}\mathbb{E} W_1^2=\frac{1}{2}$$ so that in total we have that: $$\sigma=\frac{1}{4}t^2$$