In the steady state, capacitor offers infinite resistance to $DC$, so no current flows through $6 \,\Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2\, \Omega$ and $3 \Omega$ are in parallel.
$\therefore \frac{1}{2}+\frac{1}{3}=\frac{1}{R^{'}}$
or $R^{'}=\frac{6}{5} \Omega=1.2 \Omega$
$1.2\, \Omega$ resistance is in series with $2.8 \,\Omega$
$\therefore \,\,\,R^{''}=1.2+2.8=4 \,\Omega$
Current drawn from the battery,
$I=\frac{E}{R}=\frac{6}{4}=1.5 \,A$
Potential difference between points $A$ and $B$
$V=I R^{'}=1.5 \times 1.2=1.8\, V$
$\therefore $ Current through $2 \,\Omega$ resistor
$=\frac{V}{2}=\frac{1.8}{2}=0.9\,A$
