The electric flux $\Phi_E$ through a surface is given by the dot product $\Phi_E = \vec{E} \cdot \vec{A}$. Given: Electric Field $\vec{E} = 3 \times 10^3 \hat{i}$ N/C. Side of square $a = 10$ cm $= 0.1$ m. Area magnitude $A = (0.1)^2 = 10^{-2}$ m$^2$. Since the plane of the square is parallel to the yz-plane, its area vector $\vec{A}$ points in the direction of the x-axis (normal to the yz-plane). Therefore, $\vec{A} = 10^{-2} \hat{i}$ m$^2$. Calculation: $\Phi_E = (3 \times 10^3 \hat{i}) \cdot (10^{-2} \hat{i}) = 3 \times 10^1 = 30$ Nm$^2$/C. (See NCERT Physics Class 12, Exercise 1.15).