Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given the top view of a cube.
We can assume that the electric field is uniform over the four faces of the cube.
We know that the electric flux through a surface is given by
$$\Phi_e=\vec E\cdot \vec A=EA\cos\theta$$
The area of the 4 surfaces is the same and is given by $L^2$ where $L=3$ cm.
Hence,
$$\Phi_e=EL^2\cos\theta$$
As we see in the figure below, we found the angles between the area vectors of the 4 sides and the electric field.
At side 1,
$$\Phi_1=EL^2\cos\theta_1=(500)(0.03)^2\cos150^\circ$$
$$\Phi_1=\color{red}{\bf -0.39}\;\rm N\cdot m^2/C $$
At side 2,
$$\Phi_2=EL^2\cos\theta_2=(500)(0.03)^2\cos60^\circ$$
$$\Phi_2=\color{red}{\bf 0.225}\;\rm N\cdot m^2/C $$
At side 3,
$$\Phi_3=EL^2\cos\theta_3=(500)(0.03)^2\cos30^\circ$$
$$\Phi_3=\color{red}{\bf 0.39}\;\rm N\cdot m^2/C $$
At side 4,
$$\Phi_4=EL^2\cos\theta_4=(500)(0.03)^2\cos60^\circ$$
$$\Phi_4=\color{red}{\bf -0.225}\;\rm N\cdot m^2/C $$
$$\color{blue}{\bf [b]}$$
The net flux through the four sides is given by
$$\Phi_{net}=\Phi_1+\Phi_2+\Phi_3+\Phi_4$$
$$\Phi_{net}=-0.39+0.225+0.39-0.225$$
$$\Phi_{net}=\color{red}{\bf 0}\;\rm N\cdot m^2/C $$
