Answer
$$\dot{\psi}=2.35 \mathrm{rev} / \mathrm{h}
$$
Work Step by Step
$$
I=1600(1.8)^2, \quad I_z=1600(1.2)^2
$$
$$
\begin{aligned}
& \tan \theta=\left(\frac{I}{I_z}\right) \tan \beta \\
& \tan 20^{\circ}=\left(\frac{1600(1.8)^2}{1600(1.2)^2}\right) \tan \beta \\
& \beta=9.189^{\circ}
\end{aligned}
$$
Using the law of sines:
$$
\begin{aligned}
& \frac{\sin 9.189^{\circ}}{2}=\frac{\sin \left(20^{\circ}-9.189^{\circ}\right)}{\dot{\psi}} \\
& \dot{\psi}=2.35 \mathrm{rev} / \mathrm{h}
\end{aligned}
$$
