In the given circuit, the current through 2   Ω resistor is

At the steady state, the capacitor behaves as open circuit. The
figure below represents the simple circuit,

Apply Kirchhoff law in loop I,

$-6+3\left(I_1+I_2\right)+2.8{\mathrm{I}}_1=0$

$5.8{\mathrm{I}}_1-3{\mathrm{I}}_2=6 ....\left(1\right)$

Apply Kirchhoff law in loop II,

$2{\mathrm{I}}_2+3\left(I_2-I_1\right)=0 ....\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$

$I_2=0.9 \mathrm{A}$

Thus, the current in $2 \mathrm{\Omega }$ resistor is $0.9 \mathrm{A}.$