A thin uniform shell has a mass of 30 kg and a radius of 0.60 m…

Step 1. (a) Inside. Field $0$; potential $-6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2 \times \dfrac{30\ \text{kg}}{0.60\ \text{m}} = -3.34 \times 10^{-9}\ \text{J/kg}$.

Step 2. (b) Surface. Field $\dfrac{GM}{a^2} = 5.56 \times 10^{-9}\ \text{N/kg}$; potential $-3.34 \times 10^{-9}\ \text{J/kg}$.

Step 3. (c) Outside. Field $\dfrac{GM}{r^2} = 8.89 \times 10^{-10}\ \text{N/kg}$; potential $-\dfrac{GM}{r} = -1.33 \times 10^{-9}\ \text{J/kg}$.