Correct Answer - A
Let us refer to the coil as circuit 1 and the soleniod circuite 2. The field in the central region of the solenoid is uniform, so the flux through the coil is
`phi_(12)=B_(A)A_(1)=(mu_(0)n_(2)I_(2))A_(1)`
where `n_(2)=N_(2)//l=1500` turn/m
The mutual inductance si
`M=(N_(2)phi_(12))/(I_(2))=mu_(0)n_(2)N_(1)A_(1)`
`=(4pixx10^(-7)Tm A^(-1))(1500m^(-1))(10)(4xx10^(-4)m^(2))`
`7.54xx10^(-6)H=7.54 muH`