Two capacitors with capacitances of 4.0 and 6.0 are connected in...

The equivalent capacitance $C_{eq}$ of the two capacitors connected in series is given by:

$C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{4.0 \times 6.0}{4.0 + 6.0} = 2.4 \mu\text{F}$

The total charge $Q$ drawn from the battery is:

$Q = C_{eq} V = 2.4 \mu\text{F} \times 20 \text{ V} = 48 \mu\text{C}$

The energy supplied by the battery is the work done in moving this charge across the potential difference $V$:

$E = Q V = 48 \mu\text{C} \times 20 \text{ V} = 960 \mu\text{J}$