Answer
a) $ 6.56\times 10^{15}\;\rm Hz$
b) $1.05 \;\rm mA$
Work Step by Step
$$\color{blue}{\bf [a]}$$
According to Newton's second law, we know that the electric force exerted on the electron is given by
$$F=m_ea_r$$
Hence,
$$\dfrac{k_ee^2}{r^2}=m_e\dfrac{v^2}{r}$$
$$\dfrac{k_ee^2}{r }=m_e v^2 \tag 1$$
where $v=2\pi r /T$ where $T$ is the periodic time for one full circle around the proton.
Recall that the frequency is reciprocal to the period $f=1/T$, so $v=2\pi r f$,
Plug into (1);
$$\dfrac{k_ee^2}{r }=m_e (2\pi r f)^2 $$
Solving for $f$;
$$f=\sqrt{\dfrac{k_ee^2}{4\pi^2 m_e r^3}}$$
Plug the known;
$$f=\sqrt{\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{4\pi^2 (9.11\times 10^{-31})(0.053\times 10^{-9})^3}}$$
$$f=\color{red}{\bf 6.56\times 10^{15}}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
We know that the current is given by
$$I=\dfrac{Q}{t}$$
So the effective current of the electron is given by the charge of the electron over the time of one full circle around the proton in the hydrogen atom.
$$I=\dfrac{e}{T}$$
where $T=1/f$, so
$$I=ef$$
Plug the known;
$$I=(1.6\times 10^{-19})(6.56\times 10^{15})$$
$$I=\color{red}{\bf 1.05}\;\rm mA$$
