A network of four capacitors capacity equal C_1=C, C_2=2 C, C_3=3 C...

Here $C_1, C_2 and C_3$ are in series.
$\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}$$\begin{aligned} \frac{1}{C^{\prime}}& =\frac{6+3+2}{6 C}=\frac{11}{6 C}\\ \Rightarrow \quad C & =\frac{6 C}{11} \end{aligned}$$All the$capacitors in branch number 1 is in series so the charge on each number capacitor is: Q^{\prime}=$\frac{6}{11}$C V Also charge on capacitor $C_4$ is $Q=4$ V \therefore $\text{ Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 \mathrm{CV}}{11 \times 4 C V}=\frac{3}{22}$