1. Energy of electron in Bohr's first orbit

For hydrogen atom, the total energy in the first Bohr orbit is $E_1=-R_H=-2.18\times 10^{-18}\,\text{J}$

The kinetic energy of the electron in Bohr orbit is $K=-E_1=2.18\times 10^{-18}\,\text{J}$

2. Relate de-Broglie wave frequency to electron kinetic energy

For the matter wave associated with the electron, $\nu=\frac{E}{h}$ Here the relevant energy of the moving electron is its kinetic energy in the orbit.

So, $\nu=\frac{2.18\times 10^{-18}}{6.6\times 10^{-34}}$

3. Calculate

$\nu=\frac{2.18}{6.6}\times 10^{16}$ $\nu\approx 0.3303\times 10^{16}$ $\nu\approx 3.303\times 10^{15}\,\text{Hz}$

4. Express in the required form

We need $\nu = (\text{number})\times 10^{13}\,\text{Hz}$

Thus, $3.303\times 10^{15}=330.3\times 10^{13}$

Nearest integer $=330$.

5. Comparison with stored answer

My derived answer is 330, but the stored correct answer is 658.

A value near $658\times 10^{13}=6.58\times 10^{15}\,\text{Hz}$ would come from using $\nu=\frac{2K}{h}=\frac{m v^2}{h}$ which is not the standard way to assign de-Broglie matter-wave frequency when using the given Bohr orbit kinetic energy. Using the provided data and the usual relation $\nu=E/h$ for the electron's kinetic energy in the orbit gives 330.