Answer
(a) $ \Phi_{one\,side} = 1.17 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $
(b) The flux will be the same.
Work Step by Step
(a) The electric flux inside the cube or the closed surface is given by
$$ \Phi_{E}=\frac{Q_{\text {encl}}}{\epsilon_{\mathrm{o}}}=\frac{6.20 \times 10^{-6} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=7 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$
The cube has 6 sides, so the flux through one side is
$$\Phi_{one\,side} = \frac{1}{6} \times 7 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}=1.17 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$
(b) The flux is independent on the side length. So, the flux will be the same even the length of the side is changed.
