In steady state, current through capacitor branch is zero.

$\begin{array}{l}{\mathrm{R}}_{eq}=\frac{2\times 3}{2+3}+2.8=4\Omega ; \mathrm{i}=\frac{6}{4}=1.5\mathrm{A}\\ {\mathrm{V}}_{2.8\mathrm{\Omega }}=2.8\times 1.5=4.2\mathrm{V};{\mathrm{V}}_{\mathrm{C}}=6-{\mathrm{V}}_{2.8\mathrm{\Omega }}=1.8\mathrm{V}\\ \mathrm{q}={\mathrm{CV}}_{\mathrm{C}}=0.2\mathrm{\mu F}\times 1.8V=0.36\mathrm{\mu C}\end{array}$