(D) In a steady state,the capacitor branch acts as an open circuit,so no current flows through the branch containing the capacitor $C$.
The circuit consists of a $12 \, V$ battery in series with a $2 \, \Omega$ resistor and a $6 \, \Omega$ resistor.
Total resistance of the circuit $R_{eq} = 2 \, \Omega + 6 \, \Omega = 8 \, \Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{12 \, V}{8 \, \Omega} = 1.5 \, A$.
The potential difference across the capacitor $C$ is equal to the potential difference across the $6 \, \Omega$ resistor because they are connected in parallel to the same nodes.
Potential difference $V_C = I \times 6 \, \Omega = 1.5 \, A \times 6 \, \Omega = 9 \, V$.
The charge on the capacitor is $Q = C \times V_C = 2 \, \mu F \times 9 \, V = 18 \, \mu C$.