(a) The plane of the square is parallel to the $y-z$ plane. Hence, angle between the unit vector normal to the plane and electric field, $\theta=0^{\circ}$ Flux (ϕ) through the plane is given by the relation, $\phi\;=|\vec{E}| A \cos \theta$ $\;=2\times10^{3}\times0.04\times\cos0^{\circ}$ $\;=80\,\text{N}\,\text{m}^2/\text{C}$ (b) Plane makes an angle of $30^{\circ}$ with the $x$-axis. Hence, angle between the unit vector normal to the plane and electric field, $\theta=60^{\circ}$ Flux, $\phi\;=|\vec{E}| A \cos \theta$ $\;=2\times10^{3}\times0.04\times\cos60^{\circ}$ $\;=40\,\text{N}\,\text{m}^2/\text{C}$