Answer
$171\;\rm kHz$
Work Step by Step
By exploiting the analogy with the potential energy of a mass on a spring,
$$U_{\rm proton}=\frac{1}{2}k_{\rm spring}x^2$$
where $k$ is the spring constant.
$$q_{\rm proton}V=\frac{1}{2}k_{\rm spring}x^2$$
$$6000q_{\rm proton}x^2=\frac{1}{2}k_{\rm spring}x^2$$
Hence,
$$k_{\rm spring}=12,000q_{\rm proton}\tag 1$$
The frequency of a simple harmonic motion of a spring is given by
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k_{\rm spring}}{m_{\rm proton}}}$$
Plug from (1),
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{12,000q_{\rm proton}}{m_{\rm proton}}}$$
Plug the known;
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{12,000(1.6\times 10^{-19})}{(1.67\times 10^{-27})}}$$
$$f=\color{red}{\bf 171}\;\rm kHz$$
