This is a numerical-answer question. The correct answer is shown below.
[Solution by Dr. Abubacker] $T=\frac{2\pi m}{eB}\Rightarrow B=\frac{2\pi m}{eT}=\frac{2\pi\times9\times10^{-31}}{1.6\times10^{-19}\times75\times10^{-9}}=4.71\times10^{-4}$ T. $B=\mu_0nI\Rightarrow n=\frac{4.71\times10^{-4}}{4\pi\times10^{-7}\times1.5}=250$ turns/m. Answer: 250.