If the charge is placed at the centre of one side, flux through the cube is $\frac{\mathrm{q}}{{\mathrm{m\epsilon }}_0}$where ‘n’ is

As the charge is placed at the center of the side (edge center), one-fourth of the charge will lie inside the cube.

As the total flux coming out of a charge is given by $\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$

The flux through the given cube will be given by $\mathrm{\varphi }=\frac{\mathrm{q}}{4{\mathrm{ϵ}}_0}\Rightarrow \mathrm{n}=4$