An 8.0 µF air capacitor is charged to 50 V…

Step 1. Battery connected. The voltage stays $50\ \text{V}$. The charge becomes $Q' = 4.0\ \mu\text{F} \times 50\ \text{V} = 200\ \mu\text{C}$, half of the old value. The energy is $\tfrac{1}{2} \times 4.0\ \mu\text{F} \times (50\ \text{V})^2 = 5.0\ \text{mJ}$.

Step 2. Battery removed. The charge stays $400\ \mu\text{C}$. The voltage becomes $V' = \dfrac{400\ \mu\text{C}}{4.0\ \mu\text{F}} = 100\ \text{V}$. The energy is $\dfrac{Q^2}{2C'} = \dfrac{(400\ \mu\text{C})^2}{2 \times 4.0\ \mu\text{F}} = 20\ \text{mJ}$.

Step 3. The work. With the battery removed, the only energy source is you. The energy rose from $10\ \text{mJ}$ to $20\ \text{mJ}$, so you did $10\ \text{mJ}$ of work.