To find the flux through the square, imagine the square as one face of a cube with side length 10 cm. Since the charge is 5 cm above the center of the square, it is located exactly at the center of this imaginary cube. According to Gauss's Law, the total flux through the cube is $\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$. Due to symmetry, the flux through one face (the square) is $\frac{1}{6}$ of the total flux.

Calculation: $\Phi_{\text{square}} = \frac{1}{6} \frac{q}{\varepsilon_0}$ $q = 10 \mu\text{C} = 10 \times 10^{-6} \text{ C}$ $\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}$

$\Phi_{\text{square}} = \frac{10 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}} = \frac{10 \times 10^6}{53.124} \approx 1.88 \times 10^5 \text{ Nm}^2\text{C}^{-1}$. (See NCERT Physics Class 12, Exercise 1.18).