Compute $$\mathbb{E} \left( \int_0^1 \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2.$$
Let $f(x) = x^2$, and for reference, we note that $f'(x) = 2x$ and $f''(x) = 2$. Furthermore, let $X_t$ be the process defined by $$X_t : = \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s.$$ Ito's formula gives us that $$f(X_t) - f(0) = \int_0^t f'(X_s) dX_s + \frac{1}{2} \int_0^t f''(X_s) ds.$$ Therefore, we see that \begin{eqnarray*} \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2 &=& \int_0^t 2 \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right) dX_s + \frac{1}{2} \int_0^t 2 ds \\ &=& 2\left( \int_0^t \left |B_s \right|^{\frac{1}{2}} dB_s \right) \left( X_t - X_0 \right) + t \\ &=& 2 \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2 + t \end{eqnarray*}