A capacitor with a capacitance of 100 is charged to a potential...

Initial charge on the capacitor, $Q_i = CV_i = 100 \times 10^{-6} \times 24 = 2400 \mu\text{C}$

Final charge on the capacitor, $Q_f = CV_f = 100 \times 10^{-6} \times 12 = 1200 \mu\text{C}$

Change in charge on the positive plate of the capacitor, $\Delta Q = Q_f - Q_i = 1200 \mu\text{C} - 2400 \mu\text{C} = -1200 \mu\text{C}$

Since the final charge is less than the initial charge, a positive charge of $1200 \mu\text{C}$ flows from the positive plate of the capacitor into the positive terminal of the $12 \text{ V}$ battery.

When positive charge enters the positive terminal of a battery, work is done on the battery.

Magnitude of work done on the battery $= |\Delta Q| \times V = 1200 \times 10^{-6} \times 12 = 14400 \times 10^{-6} \text{ J} = 14.4 \text{ mJ}$