Answer
$ 7.8\times 10^{13}\;\rm Hz$
Work Step by Step
We know that the frequency for a simple harmonic motion with a spring is given by
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}\tag 1$$
where $m$ here is the mass of the hydrogen atm the author told us since it is the one that oscillates back and forth.
So the only unknown here is the spring constant between the two atoms.
We know that the elastic potential energy is given by
$$U_s=\frac{1}{2}k(\Delta x)^2$$
where $(\Delta x)$ here is the stretched or compressed distance of the spring from its equilibrium point,
So,
$$k=\dfrac{2U_s}{(\Delta x)^2}\tag 2$$
In the given graph we can see that the equilibrium point is at $x_0=0.13\;\rm nm$
So, at $x=0.13$ nm, $U_s=0$ J.
Now we need to take the data of three or more points to calculate the spring constant or its average.
$\bullet$ At $x=0.085\;\rm nm$, $\Delta x=0.13-0.085= 0.045\;\rm nm$, and $U_s=4\times 10^{-19}\;\rm J$.
So, $$k=\dfrac{2(4\times 10^{-19})}{(0.045\times 10^{-9})^2}=\bf 395.1\;\rm N/m$$
$\bullet$ At $x=0.10\;\rm nm$, $\Delta x=0.13-0.10= 0.03\;\rm nm$, and $U_s=1.8\times 10^{-19}\;\rm J$.
So, $$k=\dfrac{2(1.8\times 10^{-19})}{(0.03\times 10^{-9})^2}=\bf 400\;\rm N/m$$
$\bullet$ At $x=0.15\;\rm nm$, $|\Delta x|=0.13-0.15= 0.02\;\rm nm$, and $U_s=0.8\times 10^{-19}\;\rm J$.
So, $$k=\dfrac{2(0.8\times 10^{-19})}{(0.02\times 10^{-9})^2}=\bf 400\;\rm N/m$$
$\bullet$ At $x=0.16\;\rm nm$, $|\Delta x|=0.13-0.16= 0.03\;\rm nm$, and $U_s=1.8\times 10^{-19}\;\rm J$.
So, $$k=\dfrac{2(1.8\times 10^{-19})}{(0.03\times 10^{-9})^2}=\bf 400\;\rm N/m$$
From these 4 results above, it is obvious that $k\approx 400\;\rm N/m$.
Plug into (1) and plug in the hydrogen mass.
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{400}{(1.67 \times 10^{-27})}} $$
$$f=\color{red}{\bf 7.8\times 10^{13}}\;\rm Hz$$
