(A) Net outward flux through the surface of the box,$\phi = 8.0 \times 10^{3} \; N m^{2} / C$.
According to Gauss's Law,the flux $\phi$ is related to the net charge $q$ enclosed by the surface as $\phi = \frac{q}{\varepsilon_{0}}$.
Here,$\varepsilon_{0} = 8.854 \times 10^{-12} \; N^{-1} C^{2} m^{-2}$ is the permittivity of free space.
Therefore,$q = \varepsilon_{0} \phi = (8.854 \times 10^{-12}) \times (8.0 \times 10^{3}) \; C$.
$q = 7.0832 \times 10^{-8} \; C \approx 0.07 \; \mu C$.
Thus,the net charge inside the box is $0.07 \; \mu C$.
$(b)$ No.
If the net outward flux through the surface is zero,it implies that the net charge enclosed by the surface is zero (since $\phi = q_{net} / \varepsilon_{0}$). This does not mean there are no charges inside; it only means that the total algebraic sum of all charges inside the box is zero. The box could contain equal amounts of positive and negative charges.