= Solution
For a long air-cored <solenoid>, <Ampère's law> applied to a rectangular loop running inside and outside the winding gives $B\ell\simeq\mu_0NI$, with the external <magnetic field> negligible away from the ends. Hence $B=\mu_0nI$ along the winding axis, where $n=N/\ell$. Each turn links <magnetic flux> $BA$ with $A=\pi a^2$, so the total <magnetic flux linkage> is $NBA=LI$. It follows that
$$
\boxed{B=\mu_0\frac N\ell I,\qquad
L=\mu_0\frac{N^2\pi a^2}{\ell}=\mu_0n^2\pi a^2\ell.}
$$
Here $n=2000\,\mathrm{m^{-1}}$, $a=0.010\,\mathrm m$ and $\ell=0.50\,\mathrm m$, so $N=1000$ and
$$
\boxed{L=7.90\times10^{-4}\,\mathrm H=0.790\,\mathrm{mH}.}
$$
The value uses the stated long-solenoid approximation and ignores end corrections and winding resistance.
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