In this figure the resistance of the coil of galvanometer G is 2 Ω...

Current flowing through both capacitors will be zero at steady state.

The potential difference across capacitor ${C}_{1}$ will be the sum of the potential difference across the resistor $6Ω$ and the galvanometer resistance. Mathematically,

${V}_{{C}_{1}} = i\left[6 Ω + {R}_{G}\right] . . . \left(1\right)$

Similarly, the potential difference across capacitor ${C}_{2}$ will be the sum of the potential difference across the resistor $8Ω$ and the galvanometer resistance. Mathematically,

${V}_{{C}_{2}} = i\left[{R}_{G} + 8 Ω\right] . . . \left(2\right)$

Divide equation (1) by equation (2) to obtain the required ratio.

$\begin{cases}\dfrac{{V}_{{C}_{1}}}{{V}_{{C}_{2}}} & = & \dfrac{i\left[6 Ω + {R}_{G}\right]}{i\left[{R}_{G} + 8Ω\right]} \\ & = & \dfrac{6 Ω + 2Ω}{2Ω + 8Ω} \\ & = & \dfrac{4}{5}\end{cases}$