In steady state current through the branch having capacitor is zero.
$ \therefore $ $ \frac{1}{R}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} $
$ \frac{1}{R}=\frac{6+3+2}{6} $
$ R=\frac{6}{11} $
As $ V=iR $
$ \therefore $ $ 6=i\times \frac{6}{11} $
Current through the battery $ i=11\text{ }A $
Charge on the capacitor $ q=CV $
$ \Rightarrow $ $ q=0.5\times {{10}^{-6}}\times 6 $
$ q=3\mu C $