Answer
$2.233 \times10^{-5} T\cdot m/A $
Work Step by Step
Use equation 20–8 for the magnetic field inside a solenoid, but replace $\mu_o$ with the permeability of iron.
$$B=\frac{\mu NI}{\mathcal{l}}$$
$$\mu=\frac{B\mathcal{l}}{NI}$$
$$\mu=\frac{(2.2T)(0.38m)}{(780)(48A)}$$
$$=2.233 \times10^{-5} T\cdot m/A \approx2.2 \times10^{-5} T\cdot m/A\approx 18\mu_o$$