b
In the steady state, no current is passing through capacitor. Let the charge on each capacitor be \(q\). since, the current through galvanometer is zero.

\(\therefore I_{1}=I_{2}\)

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

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

\(\therefore \frac{C_{1}}{C_{2}}=\frac{R_{2}}{R_{1}}\)