Gauss law

Note that the charge is located on the surface of the cube and not in its inner volume, making it improper to use Stokes theorem. We note C1 the cube, and C2 the C1 "mirror" cube relative to the charge (C1 and C2 both have the charge at the center of their common face).

The flux of electric field through C2 is equal to the one through of C1 from symmetry, and both are equal to half the flux through C1UC2.

But now the particle is in the volume (C1UC2) and by Stokes theorem (or Gauss theorem), the flux of eletrical field equals the total charge in the volume divided by permitivity of vacuum:

$flux(E/C1 U C2) = \frac{2 \times q}{e0}$

And then:

$flux(E/C1) = \frac{flux(E/C1 U C2)}{2} = \frac{q}{e0}$

I have a similar solution to Mat Baluch :

Construct a Gaussian Cuboid made of the $2$ original Gaussian Cubes with the point charge centered right between the $2$ cubes and right at the geometric centre of the Gaussian cuboid.Now Flux through entire cuboid is $2Q/e_0$ from Gauss Law and since $q_{enc} = 2q$ .But flux through each of the cubes is just its half that is $q/e_0$ .