Answer
$f = 3.5 \mathrm{Hz}$
Work Step by Step
The target is to calculate the frequency $f$ and from the next equation we have a relation between the ferquency and the period by
$$f=\frac{1}{T}=\frac{1}{2 \pi \sqrt{L / a}}$$
Where $a$ is the acceleration. The motion here is centrifugal, so this acceleration is related to the radius of the circular motion by
$$a = (v^2/R)$$
Substitute by this expression of $a$ into the frequency equation to get $f$ by
\begin{aligned}
f =\frac{v}{2 \pi} \sqrt{1 / L R} =\frac{70 \mathrm{m} / \mathrm{s}}{2 \pi} \sqrt{1 /(0.2 \mathrm{m})(50 \mathrm{m})} =3.5 \mathrm{Hz}
\end{aligned}
