A dipole is made of +1.5 µC and -1.5 µC, joined by a light rod 2.0 cm long…

Step 1. The moment. $p = q \times 2a = 1.5 \times 10^{-6}\ \text{C} \times 0.020\ \text{m} = 3.0 \times 10^{-8}\ \text{C m}$.

Step 2. The product $pE$. $pE = 3.0 \times 10^{-8}\ \text{C m} \times 4.0 \times 10^{4}\ \text{N/C} = 1.20 \times 10^{-3}\ \text{N m}$. This is the greatest torque, at $90^\circ$: $1.20$ mN m.

Step 3. At $40^\circ$. $\tau = pE\sin 40^\circ = 1.20 \times 10^{-3} \times 0.643\ \text{N m} = 7.71 \times 10^{-4}\ \text{N m}$, that is $0.771$ mN m.

Step 4. The work. $W = pE(\cos 0^\circ - \cos 110^\circ) = 1.20 \times 10^{-3}\ \text{J} \times (1 + 0.342) = 1.61 \times 10^{-3}\ \text{J}$, that is $1.61$ mJ.