In the network shown below, the charge accumulated in the capacitor...

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

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

The current flowing through the circuit is

$I = \dfrac{3V}{10Ω}$

Potential difference on $6Ω$ resistor

$V ' = \dfrac{3}{10} \times 6 = 1 . 8V$

The charge on the capacitor is,

$Q = CV ' = 4\mu F \times 1 . 8V = 7 . 2\mu C$