(a) The plane of the square is parallel to the $\mathit{y}-\mathit{z}$ plane. Hence, angle between the unit vector normal to the plane and electric field, $\mathit{\theta }=0^{\circ }$
Flux (ϕ) through the plane is given by the relation,
$\mathit{\varphi }=|\overrightarrow{\mathit{E}}|\mathit{A}\cos \mathit{\theta }$
$=4\times {10}^3\times 0.25\times \cos 0^{\circ }$
$=10\mathrm{N}{\mathrm{m}}^2∕\mathrm{C}$
(b) Plane makes an angle of ${30}^{\circ }$ with the $\mathit{x}$-axis. Hence, angle between the unit vector normal to the plane and electric field, $\mathit{\theta }={60}^{\circ }$
Flux, $\mathit{\varphi }=|\overrightarrow{\mathit{E}}|\mathit{A}\cos \mathit{\theta }$
$=4\times {10}^3\times 0.25\times \cos {60}^{\circ }$
$=5\mathrm{N}{\mathrm{m}}^2∕\mathrm{C}$