If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ then charge enclosed…
If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ then charge enclosed in the surface is ( $\varepsilon_0=$ permittivity of free space)
$\frac{\phi_2-\phi_1}{\varepsilon_0}$
$\frac{\phi_2+\phi_1}{\varepsilon_0}$
$\frac{\phi_1-\phi_2}{\varepsilon_0}$
$\varepsilon_0\left(\phi_2-\phi_1\right)$
Solution
$\begin{aligned}
& \Phi=\frac{q_{i n}}{\varepsilon_o} \\
& q_{i n}=\varepsilon_o \Phi \\
& q_{\text {in }}=\left(\Phi_2-\Phi_1\right) \varepsilon_o
\end{aligned}$ Hence, option D is the correct answer.