We use the formula for the number of turns per unit length of a solenoid:
$\mathit{L}={\mathit{\mu }}_0{\mathit{n}}^2\mathit{A}\mathit{l}$
where:
L = 0.016H (Inductance)
l = 0.81m (Length of solenoid)
r = 0.02m (Radius)
$\mathit{A}=\mathit{\pi }{\mathit{r}}^2$ (Cross-sectional area)
n = $\frac{\mathit{N}}{\mathit{l}}$ (Turns per unit length)
Rearranging for N:
$\mathit{N}=\sqrt{\frac{\mathit{L}}{{\mathit{\mu }}_0\mathit{A}}}\times \mathit{l}$
Substituting values:
$\mathit{N}=\sqrt{\frac{0.016}{\left(4\mathit{\pi }\times {10}^{-7}\right)\times \left(\mathit{\pi }\times {\left(0.02\right)}^2\right)}\times 0.81}$ $=\sqrt{\frac{16\times 81}{100\times 1000\times {10}^{-11}\times 16{\mathit{\pi }}^2}}$
$\mathit{N}=\frac{4\times 9\times 1000}{4\mathit{\pi }}=\frac{9000}{\mathit{\pi }}$
$=\mathit{N}\approx 2866$