A charge ‘q’ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\overrightarrow{E}$ through the shaded area is:

Imagine 7 more such small cubes and arrange all 8 in such a way that charge q is at the center of big cube.

Flux through big cube $=\frac{q}{{\in }_0}$

Thus, flux through small cube $=\frac{q}{8{\in }_0}$

Now , the flux through the 3 adjoining faces to origin of small cube is zero, because field lines does not cross.

So, we can say,  flux through 3 opposite faces $=\frac{q}{8{\in }_0}$  
Hence, flux through two half faces = flux through one full face = $\varphi =\frac{q}{24{\in }_0}$