A capacitor of capacitance 4 is connected in series with another...

Equivalent capacitance of the series combination of $4 \mu\text{F}$ and $12 \mu\text{F}$ is:
$C_s = \dfrac{4 \times 12}{4 + 12} = \dfrac{48}{16} = 3 \mu\text{F}$

The charge on the $4 \mu\text{F}$ capacitor is the same as the charge on this series branch, since capacitors in series carry the same charge.
The voltage across the series combination is the battery voltage, $20 \text{ V}$.
$Q = C_s \times V = 3 \mu\text{F} \times 20 \text{ V} = 60 \mu\text{C}$

The total equivalent capacitance of the circuit is the parallel combination of $C_s$ and the $5 \mu\text{F}$ capacitor:
$C_{eq} = 3 \mu\text{F} + 5 \mu\text{F} = 8 \mu\text{F}$

The total electrostatic energy stored in the circuit is:
$U = \dfrac{1}{2} C_{eq} V^2 = \dfrac{1}{2} \times (8 \times 10^{-6} \text{ F}) \times (20 \text{ V})^2$
$U = 4 \times 10^{-6} \times 400 = 1600 \times 10^{-6} \text{ J} = 1.6 \text{ mJ}$

Answer: $60 \mu\text{C}, 1.6 \text{ mJ}$