The capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected in series. Let the resultant of these ca[acotors is $C^{\prime}$

$\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}$

$=\frac{6+3+2}{6 C}=\frac{11}{6 C}$

$\Rightarrow C^{\prime}=\frac{6 C}{11}$

Now $\mathrm{C}^{\prime}$ and $C_{4}$ from parallel combination giving $C^{\prime \prime}=C^{\prime}+C_{4}$

$=\frac{6 C}{11}+4 C=\frac{50 C}{11}$

Net charge $q=C^{\prime \prime} V$

$=\frac{50}{11} \mathrm{CV}$

Total charge flowing through $C_{1}, C_{2}$ and $C_{3}$ will be

$q^{\prime}=q^{\prime}-q_4$

$=\frac{50}{11} C V-4 C V=\frac{6 C V}{11}$

since, $C_{1}, C_{2}$ and $C_{3}$ are in series combination hence, charge flowing through these will be the same. Hence,

$q_{2}=q_{1}=q_{3}=q^{\prime}=\frac{6 C V}{11}$

Thus, $\frac{q_{2}}{q_{4}}=\frac{6 C V / 11}{4 C V}=\frac{3}{22}$