In the network shown below the charge accumulated in the capacitor in steady state will be:

at steady state capacitor becomes open circuit
 

Voltage across capacitor = voltage across AB ${\mathrm{V}}_{\mathrm{AB}}=\frac{3\times 6}{6+4}=\frac{18}{10}=1.8\mathrm{V}$
So, charge on capacitor = 4 x 1.8 μC = 7.2 μC