A 40 µF capacitor in a medical unit must store 80 J…

Step 1. The voltage. From $U = \tfrac{1}{2}CV^2$ we want $V$, so $V = \sqrt{\dfrac{2U}{C}} = \sqrt{\dfrac{2 \times 80\ \text{J}}{40 \times 10^{-6}\ \text{F}}} = \sqrt{4.0 \times 10^6\ \text{V}^2} = 2000\ \text{V}$.

Step 2. The charge. $Q = CV = 40 \times 10^{-6}\ \text{F} \times 2000\ \text{V} = 0.080\ \text{C} = 80\ \text{mC}$.

Step 3. Half the voltage. $U \propto V^2$, so half the voltage gives a quarter of the energy: $\dfrac{80\ \text{J}}{4} = 20\ \text{J}$.