
Capacitance of each capacitor
C1=21A3ϵ0=6Aϵ0 and C2=1A4ϵ0=4Aϵ0
Here, A is area of plate.
Equivalent capacitance is
Ceq=C1+C2C1C2=6Aϵ0+4Aϵ06Aϵ0×4Aϵ0=1024Aϵ0
Now, net charge stored is qnet=Ceq(ΔV)=1024Aϵ0(100)=240Aϵ0
Potential drop across C2 is ΔV2=4Aϵ0240Aϵ0=60V
Thus, voltage of the conducting foil is Vfoil=60V