Answer
$1.13\times 10^{14}\;\rm Hz$
Work Step by Step
The electric force between the electron and the positron is given by
$$F=\dfrac{kq^2}{r^2}$$
We can ignore the signs of the charges since we know that the force between them is an attractive electric force.
Now let's work with one of them, the net force exerted on the positron is given by
$$\sum F=\dfrac{kq^2}{r^2}=ma_r$$
where $a_r=v^2/(0.5r)$ where the radius of rotation is half the distance between the two particles.
$$ \dfrac{kq^2}{r^2}=\dfrac{2mv^2}{r}$$
Hence,
$$v^2=\dfrac{kq^2}{2mr}$$
where $v=2\pi (0.5r)/T= \pi r f$;
$$ \pi^2 r^2 f^2=\dfrac{kq^2}{2mr}$$
$$ f =\sqrt{\dfrac{kq^2}{ 2\pi^2mr^3}}$$
Plug the known
$$ f =\sqrt{\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{ 2\pi^2(9.11\times 10^{-31})(1\times 10^{-9})^3}}$$
$$f=\color{red}{\bf 1.13\times 10^{14}}\;\rm Hz$$
