Answer
$0.17\frac{rad}{s}$
Work Step by Step
We know that
$\Delta t=\frac{\Delta t_{\circ}}{\sqrt{1-v^2/c^2}}$
$\implies \frac{2\pi}{\omega}=\frac{\frac{2\pi}{\omega_{\circ}}}{\sqrt{1-v^2/c^2}}$
This simplifies to:
$\omega=\omega_{\circ}\sqrt{1-v^2/c^2}$
We plug in the known values to obtain:
$\omega=0.29\times \sqrt{1-(0.82)^2}=0.17\frac{rad}{s}$
