A circular coil has 50 turns of radius 3.5 cm…

Step 1. The moment. $A = \pi r^2 = \pi \times (0.035\ \text{m})^2 = 3.85 \times 10^{-3}\ \text{m}^2$. So $m = NIA = 50 \times 1.8\ \text{A} \times 3.85 \times 10^{-3}\ \text{m}^2 = 0.346\ \text{A m}^2$.

Step 2. The energy scale. $mB = 0.346\ \text{A m}^2 \times 0.42\ \text{T} = 0.145\ \text{J}$. At the start, $U = -mB$.

Step 3. A quarter turn. At $90^\circ$, $U = 0$. The work is $W = 0 - (-mB) = mB = 0.145\ \text{J}$.

Step 4. A half turn. At $180^\circ$, $U = +mB$. The work is $W = mB - (-mB) = 2mB = 0.291\ \text{J}$.