No current will flow in capacitor in steady state, current flowing in the circuit in steady state $$ I=\frac{3}{6+4}=\frac{3}{10} $$ Potential difference on $6 \Omega$ resistance $$ \mathrm{V}=6 \times \frac{3}{10}=1.8 \mathrm{volt} $$ Capacitor will have same potential so charge, $$ \mathrm{q}=\mathrm{CV}=(4 \mu \mathrm{~F}) .(1.8 \mathrm{volt})=7.2 \mu \mathrm{C} . $$