(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 \; = \left| \vec{E} \right| A \cos \theta$$\; = 4 \times 10^3 \times 0.25 \times \cos 0^{\circ}$$\; = 10 \text{ N 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 \; = \left| \vec{E} \right| A \cos \theta$$\; = 4 \times 10^3 \times 0.25 \times \cos 60^{\circ}$$\; = 5 \text{ N m}^2/\text{C}$