When switch is opened, the circuit is as following

since the capacitors are in series plates b and c will have equal and opposite charges

∴ q b + q c = 0 ; also charge on each capacitor = 18µC

when switch is closed, the circuit is as following

In steady state, the current in the resistances is 1 amp. Potential difference across 3µF

= potential difference across 3Ω = 3 volt.

Similarly p.d. across 6 µF is 6 volt

∴ charge on plate b = – 9 µC

and charge on plate c = + 36 µC

∴ charge on plates b and c = +36 – 9 = +27 µC.

The change in charges on plates b and c goes through wire from Y to X.

and Δ q = 27µC