Balls of 4.0 kg and 6.0 kg are 0.30 m apart…

Step 1. $U = -6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2 \times \dfrac{24\ \text{kg}^2}{0.30\ \text{m}} = -5.34 \times 10^{-9}\ \text{J}$.

Step 2. To reach $U = 0$: $5.34 \times 10^{-9}\ \text{J}$.

Step 3. At twice the distance, half the size: $-2.67 \times 10^{-9}\ \text{J}$.