Determine the current in 2 resistor.

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