= 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 $B_A=\mu_0n_1I/s$, approximately uniform inside and negligible outside. Therefore the flux through one cross-section, or one turn of either coil, is
$$
\boxed{\Phi=\frac{\mu_0n_1IA_t}{s}=\frac{\mu_0n_1I\pi d^2}{4s}.}
$$
The 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.
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