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

At steady state, capacitor branch carries no current.

So middle branch is open, and only top and bottom branches determine potential across capacitor.

Top branch: battery  with resistor $$4Ω$$

Bottom branch: resistor $$6Ω$$

They form one closed loop.

Current:

$$I=\frac{3}{4+6}=0.3A$$

Potential drop across bottom $$6Ω$$ resistor:

V=IR

$$=0.3(6)=1.8V$$

This is the potential difference between left and right nodes.

Since in steady state no current flows through middle $$6Ω$$, no voltage drop occurs across that resistor, so capacitor gets full node voltage:

$$V_C=1.8V$$

Charge stored:

$$Q=CV$$

$$=(4\mu F)(1.8)$$

$$=7.2μC$$