Answer
${\bf 3.5\times 10^{-4}}\;\rm Wb$
Work Step by Step
As we see from the given figure, the loop is now divided into two surfaces, the top surface (1) which is a rectangle, and the left surface (2) which is an identical rectangle as well.
Hence, the net flux through the whole loop is given by
$$\Phi_{net}=\Phi_{1}+\Phi_{2}$$
where $\Phi=AB\cos\theta$
$$\Phi_{net}=A_1B\cos45^\circ+A_2B\cos45^\circ$$
where $A_1=A_2=A$
$$\Phi_{net}=2A B\cos45^\circ$$
Plug the known;
$$\Phi_{net}=2(0.05\times 0.10)(0.05)\cos45^\circ$$
$$\Phi_{\rm net}=\color{red}{\bf 3.5\times 10^{-4}}\;\rm Wb$$
