We use the formula for the number of turns per unit length of a solenoid:$L=\mu_0 n^2 A l$where:L = 0.016H (Inductance)l = 0.81m (Length of solenoid)r = 0.02m (Radius)$A = \pi r^2$ (Cross-sectional area)n = $\;\frac{N}{l}$ (Turns per unit length)Rearranging for N:$N = \sqrt{\;\frac{L}{\mu_0 A}} \times l$Substituting values:$N = \sqrt{\;\frac{0.016}{(4\pi \times 10^{-7}) \times (\pi \times (0.02)^2)} \times 0.81}$ $= \sqrt{\;\frac{16 \times 81}{100 \times 1000 \times 10^{-11} \times 16 \pi^2}}$$N = \;\frac{4 \times 9 \times 1000}{4\pi} = \;\frac{9000}{\pi}$$= N \approx 2866$