A motor coil has 80 turns, each of area 0.005 m².

Step 1. (a) $\tau = 80 \times 0.25\ \text{T} \times 1.5\ \text{A} \times 0.005\ \text{m}^2 = 0.15\ \text{N m}$.

Step 2. (b) $N$ becomes $40$ turns and $I$ becomes $4.5\ \text{A}$: $\tau = 40 \times 0.25\ \text{T} \times 4.5\ \text{A} \times 0.005\ \text{m}^2 = 0.225\ \text{N m}$.