Answer
$M= \mu_o \frac{ N_1 N_2 A}{\mathcal{l}}$.
Work Step by Step
Use equation 21–2b to find the emf induced in the short coil. The relevant area is the common cross-sectional area, and the magnetic field B is the field created by the solenoid. The current changing in the solenoid causes the flux in the turns of wire to change, leading to an emf and their mutual inductance.
$$\epsilon=-N_2\frac{A\Delta B}{\Delta t}=- N_2\frac{A\Delta (\mu_oN_1I_1)/\mathcal{l}|}{\Delta t}$$
$$= -N_2\frac{ N_1\mu_o A}{\mathcal{l}}\frac{\Delta I_1}{\Delta t}$$
Compare this to equation 21–8a.
$$\epsilon =-M\frac{\Delta I_1}{\Delta t}$$
The mutual inductance M is the proportionality constant in the equation.
$$M= \mu_o \frac{ N_1 N_2 A}{\mathcal{l}}$$