In a circuit, a 20\ V battery is connected in series with a 2\...

At steady state, the capacitor is fully charged and acts as an open circuit, so no current flows through the branch containing the $4\ \Omega$ resistor and the $5\ \mu\text{F}$ capacitor.

The steady-state current flows only through the $2\ \Omega$ and $3\ \Omega$ resistors, which are effectively in series.

The current in the circuit is:
$I = \dfrac{V}{R_{\text{eq}}} = \dfrac{20}{2 + 3} = 4\ \text{A}$

The potential difference across the parallel section is the voltage drop across the $3\ \Omega$ resistor:
$V_{p} = I \times 3 = 4 \times 3 = 12\ \text{V}$

Since there is no current in the $4\ \Omega$ resistor, there is no voltage drop across it. The potential difference across the capacitor is equal to the potential difference across the parallel section, which is $12\ \text{V}$.

The charge stored on the capacitor is:
$Q = CV = 5\ \mu\text{F} \times 12\ \text{V} = 60\ \mu\text{C}$

Answer: $60\ \mu\text{C}$