$\begin{aligned} & 1 \times \sin 40^{\circ}=1.31 \sin \theta \\ & \Rightarrow \quad \sin \theta=\frac{0.64}{1.31} \Rightarrow \theta \simeq 30^{\circ} \\ & \Rightarrow \quad I=20 \mu \mathrm{~m} \times \cot \theta \\ & \therefore \quad N=\frac{2}{20 \times 10^{-6} \times \cot \theta} \\ & \quad=\frac{2 \times 10^6}{20 \times \sqrt{3}}=57735 \\ & N \simeq 57000\end{aligned}$