Let us refer to the coil as circuit $1$ and the solenoid as circuit $2$. The field in the central region of the solenoid is uniform, so the flux through the coil is
$\phi_{12}=B_{2}A_{1}=\left(\mu_{0}n_{2}I_{2}\right)A_{1}$
where $n_{2}=N_{2} l =1500$ turns/m
The mutual inductance is
$M=\frac{N_{1}\phi_{12}}{I_{2}}=\mu_{0}n_{2}N_{1}A_{1}$
$=\left(4\pi\times10^{-7}T\,m\,A^{-1}\right)\left(1500\,m^{-1}\right)\left(10\right)\left(4\times10^{-4}\,m^{2}\right)$
$=7.54\times10^{-6}\,H $
$=7.54\,\mu H$