Let us assume that the given square be one face of the cube of edge 10 cm. As charge of +10 μC is at a distance of 5 cm above the centre of a square, so it is enclosed by the cube. Hence by Gauss’s theorem, electric flux linked with the cube is ϕ = $\frac{q}{\varepsilon_0}$ = $\frac{10 \times 10^{-6}}{8.85 \times 10^{-12}}$ = 1.13 × $10^6$ N $\mathrm{m}^2\,\mathrm{C}^{-1}$ So, the magnitude of the electric flux through the square is $\phi_{\mathrm{sq}}$ = $\frac{\phi}{6}$ = $\frac{1.13}{6}$ × $10^6$ or $\phi_{\mathrm{sp}}$ = 1.9 × $10^5$ N $\mathrm{m}^2\mathrm{C}^{-1}$