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

In steady state, capacitor behaves as open circuit, so no current will flow.

The total resistance of the circuit is, $R=\left(6+4\right) \mathrm{\Omega }=10 \mathrm{\Omega }$

The current flowing through the circuit is

$I=\frac{3 \mathrm{V}}{10 \mathrm{\Omega }}$

Potential difference on $6 \mathrm{\Omega }$ resistor

$V\text{'}=\frac{3}{10}\times 6=1.8 \mathrm{V}$

The charge on the capacitor is,

$Q=CV\text{'}=4 \mathrm{\mu F}\times 1.8 \mathrm{V}=7.2 \mathrm{\mu C}$