$\therefore I_{1}=I_{2}$

The potential difference between ends of galvanometer will be zero.

$\therefore V_{A}-V_{B}=V_{A}-V_{D}$

$\therefore I_{1} R_{1}=\frac{q}{c_{1}} \ldots( i )$

Similarly, $V_{B}-V_{C}=V_{D}-V_{C}$

$I_{2} R_{2}=\frac{q}{c_{2}} \ldots( ii )$

On dividing equation $(i)$ by equation $(ii)$, we get

$\frac{I_{1} R_{1}}{I_{2} R_{2}}=\frac{q / C_{1}}{q / C_{2}}=\frac{C_{2}}{C_{1}}$

$\therefore \frac{C_{1}}{C_{2}}=\frac{R_{2}}{R_{1}}$