 Solution

Let $A_t=\pi d^2/4$ be the tube cross-section. A long uniformly wound solenoid with $n_1$ total turns has field $B_A={\mu }_0{n}_{1}I/s$, approximately uniform inside and negligible outside. Therefore the flux through one cross-section, or one turn of either coil, isThe linked flux of coil A is $n_1\Phi$; that of coil B is $n_2\Phi$. Distinguishing flux from flux linkage prevents an extra factor of the turn count.