The square can be considered as one face of a cube of edge $10 \,cm$ with a centre where charge $q$ is placed. According to Gauss's theorem for a cube, total electric flux is through all its six faces.

$\phi_{\text {Total}}=\frac{q}{\varepsilon_{0}}$

Hence, electric flux through one face of the cube i.e., through the square is

$\phi=\frac{\phi_{\text {Total}}}{6}=\frac{1}{6} \cdot \frac{q}{\varepsilon_{0}}$

Where, $\varepsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12}\, N ^{-1} \,C ^{2} \,m ^{-2}$

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

$=1.88 \times 10^{5} \,N \,m ^{2} \,C ^{-1}$

$\therefore \phi=\frac{1}{6} \cdot \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}$

Therefore, electric flux through the square is $1.88 \times 10^{5} \;N \;m ^{2} \,C ^{-1}$