b
In steady state there will be no current in branch of capacitor, so no voltage drop across \(\mathrm{R}_2=5 \Omega\)

\(\mathrm{I}_2=0\)

\(\mathrm{I}_1=\mathrm{I}_3=\frac{10}{4+6}=1 \mathrm{~A}\)

\(\mathrm{~V}_{\mathrm{R}_3}=\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{R}_2} \quad \mathrm{~V}_{\mathrm{R}_2}=0\)

\(\mathrm{I}_3 \mathrm{R}_3=\mathrm{V}_{\mathrm{c}}\)

\(\mathrm{V}_{\mathrm{c}}=1 \times 6=6 \mathrm{volt}\)

\(\mathrm{q}_{\mathrm{c}}=\mathrm{CV}_{\mathrm{c}}=10 \times 6=60 \mu \mathrm{C}\)