Answer
$6.32\times10^{-6}H$.
Work Step by Step
See Example 30.1, which gives the mutual inductance for this configuration of coils.
$$M=\frac{\mu_0 A N_1N_2}{l}$$
$$M=\frac{(4\pi \times10^{-7}Wb/m \cdot A)\pi(0.00200m)^2(800)(50)}{0.100m}$$
$$=6.32\times10^{-6}H $$
