At steady state the capacitor will be frilly charged and thus there will be no current in the $10\, \Omega$ resistance. So the effective circuit becomes
Net current from the $6 \,V$ battery,
$I=\frac{6}{\left(\frac{2 \times 3}{2+3}\right)+\frac{2.8}{1}}$
$=\frac{6}{1.2+2.8}=\frac{3}{2}=1.5 \,A$
Between $A$ and $B$, voltage is same in both resistances,
$2 I_{1}=3 I_{2}$ where
$I_{1}+I_{2}=I=1.5 $
$\Rightarrow 2 I_{1}=3\left(1.5-I_{1}\right)$
$\Rightarrow I_{1}=0.9\, A$
