Explanation

(b) The charge flowing through $C_{4}$ is
$q_{4} = C_{4} \times V = 4CV$
The series combination of $C_{1},C_{2}$ and $C_{3}$ given
$\frac{1}{C'} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C}$
$\frac{1}{C'} = \frac{6 + 3 + 2}{6C} = \frac{11}{6C} \Rightarrow C^{'} = \frac{6C}{11}$
Now, $C'$ and $C_{4}$ form parallel combination giving
$C^{''} = C^{'} + C_{4} = \frac{6C}{11} + 4C = \frac{50C}{11}$
Net charge $q = C^{''}V = \frac{50}{11}CV$
Total charge flowing through $C_{1},C_{2},C_{3}$ will be
$q^{'} = q - q_{4} = \frac{50}{11}CV - 4CV = \frac{6CV}{11}$
Since, $C_{1},C_{2},C_{3}$ are in series combination hence, charge flowing through these will be same
Hence, $q_{2} = q_{1} = q_{3} = q^{'} = \frac{6CV}{11}$
Thus, $\frac{q_{2}}{q_{4}} = \frac{6CV/11}{4CV} = \frac{3}{22}$