Steady state behavior of capacitor

At steady state in a DC circuit, the capacitor behaves like an open circuit.
So no current flows through the branch containing the capacitor.

Voltage across the capacitor

Therefore, the charge on the capacitor is determined from $Q = C V_C$ where $V_C$ is the potential difference across the capacitor.

Using the circuit values

From the given circuit, the potential difference across the capacitor comes out to be $V_C = 4\text{ V}$ and the capacitance is $C = 4\,\mu\text{F}$

Compute the charge

$Q = CV = (4\,\mu\text{F})(4\,\text{V}) = 16\,\mu\text{C}$

Final answer

Hence, the steady state charge on the capacitor is $\boxed{16\,\mu\text{C}}$