After a long time capacitor branch will behave like open circuit so circuit will be like,

Equivalent resistance is
R=6+(4∣∣(2+10))Ω=9Ω
and, the current through battery becomes 972=8A
From the circuit shown, current through 10Ω resistor becomes, (10+2+4)4×8=2A
Potential available for capacitor is = potential drop across 10Ω
∴ Potential available for capacitor is: 10×2=20V
∴ Charge on capacitor Q=(10μF)(20V)=200μC
