Explanation

(d) $\because$ charge 8q is placed at one corner of the cube, it can be imagined to be placed at the centre of a large cube which can be formed using 8 similar cubes and arranging them
Now 8q is at centre of the 8 cubes arranged to form a closed box
Total flux through the bigger cube
$= \frac{8q}{\varepsilon_{0}}$ [Gauss’s law]
$\Rightarrow$ Flux through one small cube $= \frac{1}{8} \times \frac{8q}{\varepsilon_{0}} = \frac{q}{\varepsilon_{0}}$