#### Answer

$\phi_{A}=\frac{4q}{\epsilon_0}$
$\phi_{B}=-\frac{4q}{\epsilon_0}$
$\phi_{C}=0$
$\phi_{D}=\frac{3q}{\epsilon_0}$
$\phi_{E}=0$

#### Work Step by Step

Flux through a body due to charges present outside is 0. So, there will be flux only due to charges present inside the body.
For surface A, charge inside is $ 4q$ while for B, the charge inside is $-4q$. For C, it is $0 $ and for surface D it is $3q$. For E, it is $0$.
1.Charge in surface A $=4q$ so, electric flux through $\phi_{A}=\frac{4q}{\epsilon_0}$
2. Charge in surface B $=-4q$ so, electric flux through $\phi_{B}=-\frac{4q}{\epsilon_0}$
3. Charge in surface C $=0$ so, electric flux through $\phi_{C}=0$
4.Charge in surface D $=3q$ so, electric flux through $\phi_{D}=\frac{3q}{\epsilon_0}$
5.Charge in surface E $=0$ so, electric flux through $\phi_{E}=0$
Hence,
$\phi_{A}=\frac{4q}{\epsilon_0}$
$\phi_{B}=-\frac{4q}{\epsilon_0}$
$\phi_{C}=0$
$\phi_{D}=\frac{3q}{\epsilon_0}$
$\phi_{E}=0$