Thank you for reporting, we will resolve it shortly
Q.
In the given figure, charge stored in capacitor in steady state will be
Solution:
In steady state current drawn from battery
$I=\frac{12}{4+2}=2\, A$
Potential drop across $A B=4 \times 2=8\, V$
This is drop across capacitor also
$q=C V=2 \times 10^{-6} \times 8=16\, \mu C$