1. Use Gauss's law idea via symmetry

A point charge is placed at a distance $6\,\text{cm}$ vertically above the center of a square of side $12\,\text{cm}$.

Since the side of the square is $12\,\text{cm}$, the distance of the charge from the square is also $6\,\text{cm}$.

This means the square can be imagined as one face of a cube of side $12\,\text{cm}$, with the charge at the center of the cube.

2. Total electric flux through the cube

By Gauss's law,

$\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$

where

$q = 12\,\mu\text{C} = 12 \times 10^{-6}\,\text{C}$

and

$\varepsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2/\text{N m}^2$

So,

$\Phi_{\text{total}} = \frac{12 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.355 \times 10^6\,\text{N m}^2/\text{C}$

3. Flux through one face of the cube

Because the charge is at the center of the cube, the flux is equally distributed among its 6 faces.

Thus,

$\Phi_{\text{square}} = \frac{1}{6}\cdot \frac{q}{\varepsilon_0}$

$\Phi_{\text{square}} = \frac{1}{6}(1.355 \times 10^6)$

$\Phi_{\text{square}} \approx 2.258 \times 10^5\,\text{N m}^2/\text{C}$

4. Match with the required format

The question asks for

$\underline{\hspace{1cm}} \times 10^3\,\text{N m}^2/\text{C}$

So,

$2.258 \times 10^5 = 225.8 \times 10^3$

Hence the required integer is

$\boxed{226}$

5. Comparison with stored answer

Stored correct answer = $226$

This matches our derived answer.