Answer
$6.56\times10^{15}\;\rm Hz$
Work Step by Step
We have an attractive electric force exerted on an electron by the proton. This is similar to the gravitational force between the Earth and the Moon.
According to Newton's second law,
$$\sum F_r=F_E=ma_r$$
where $F_E=kq_pq_e/r^2=kq^2/r^2$ and $a_r=v^2/r$
$$F_E=ma_r$$
$$\dfrac{k q^2}{r^{ \color{red}{\bf\not} 2}}=\dfrac{m_ev^2}{ \color{red}{\bf\not} r}$$
$$v^2= \dfrac{k q^2}{r m_e} \tag 1$$
We know that the circular speed is also given by $2\pi r/T$ where $T$ is the periodic time for one full rotation.
And we know that $T=1/f$ where $f$ is the frequency.
Thus, $v=2\pi rf$
Plug into (1),
$$ (2\pi rf)^2= \dfrac{k q^2}{r m_e} $$
Solving for $f$,
$$ f =\sqrt{ \dfrac{k q^2}{4\pi^2 r^3 m_e} }=\dfrac{q}{2\pi r}\sqrt{\dfrac{k}{rm_e}}$$
Plug the known;
$$ f = \dfrac{1.6\times 10^{-19}}{2\pi (0.053\times 10^{-9})}\sqrt{\dfrac{9\times 10^9}{(0.053\times 10^{-9})(9.11\times 10^{-31})}}$$
$$f=\color{red}{\bf 6.56\times10^{15}}\;\rm Hz$$
