Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The electric field exertred by the positively charged ring on the negative charge will pull it toward the center of the ring.
And since it is already accelerated toward the center it will pass it and moves in the negative $z$-direction. And then the ring will pull it back toward the center of the ring so it stops and then moved back toward the center and then pass the center toward the positive $z$-direction.
This process will be repeated forever which means that the charge will oscillate back and forth on the $z$-axis.
The net force exerted on the negative charge at the center of the ring is zero and that's make it always continue moving back and forth on the $z$-direction.
Now we can prove this mathematically. Recall that the electric field exerted by the ring on its axis is given by:
$$E=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{zQ}{(z^2+R^2)^{3/2}}$$
Therefore, the force exerted on the negative charge is given by:
$$F=|q|E$$
$$\boxed{F=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{-zqQ}{(z^2+R^2)^{3/2}}}$$
The negative sign indicates that the electric force always opposes the direction of the electric field, meaning the force is always directed toward the center of the ring.
$$\color{blue}{\bf [b]}$$
when $z\lt \lt R$, the force exerted by the ring on the negative charge is given by
$$F=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{-zqQ}{R^3\left(\dfrac{z^2}{R^2}+1\right)^{3/2}}$$
where $z^2/R62\approx 0$, and hence $\left(\dfrac{z^2}{R^2}+1\right)^{3/2}=1$,
$$F=-\left[\dfrac{1}{(4\pi \epsilon_0)}\dfrac{ qQ}{R^3 }\right]z$$
We cans ee that this formula is similar to Hooke's law $F=-kx$ since the term between parentheses is constant.
Hence,
$$k=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{ qQ}{R^3 }\tag 1$$
Recall that the frequency of a simple harmonic motion of a particle that obeys Hooke’s law undergoes is given by
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$
Plug from (1),
$$\boxed{f=\dfrac{1}{2\pi}\sqrt{\dfrac{1}{(4\pi \epsilon_0)}\dfrac{ qQ}{mR^3 }}}$$
$$\color{blue}{\bf [c]}$$
Plugging the known into the previous boxed formula above,
$$ f=\dfrac{1}{2\pi}\sqrt{(9\times 10^9)\dfrac{ (1.6\times 10^{-19})(1\times 10^{-13})}{(9.11\times 10^{-31})(1\times 10^{-6})^3 }} $$
$$f=\color{red}{\bf 2\times 10^{12}}\;\rm Hz$$
