How to calculate atomic number of an atom when wavelength is known?
Wikipedia (Bohr Model > Rydberg formula) tells us:


  The energy of a photon emitted by a hydrogen atom is given by the difference of two hydrogen energy levels:
  $$E=E_i-E_f=R_\mathrm{E} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$$
  where $n_f$ is the final energy level, and $n_i$ is the initial energy level.
  
  Since the energy of a photon is
  $$E=\frac{hc}{\lambda}$$
  the wavelength of the photon given off is given by
  $$\frac{1}{\lambda}=R_\infty \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right).$$
  
  This is known as the Rydberg formula, and the Rydberg constant $R$ is $R_\mathrm{E}/hc$, or $R_\mathrm{E}/2\pi$ in natural units.


Wikipedia also tells that it can be extended with $Z^2$ for hydrogen-like elements:


  The formula above can be extended for use with any hydrogen-like chemical elements with
  $$\frac{1}{\lambda_{\mathrm{vac}}} = R_\infty Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  ... where $\lambda_{\mathrm{vac}}$ is the wavelength of the light emitted in vacuum, $R_\infty$ is the Rydberg constant for this element, $Z$ is the atomic number, i.e. the number of protons in the atomic nucleus of this element and, $n_1$ and $n_2$ are integers such that $n_1 &lt; n_2$, corresponding to the principal quantum numbers of the orbitals occupied before and after.




Now you can see, that your first equation can be derived from your second equation. You also can see, that you a) forgot the squares in your first equation and b) made a sign error.

$\lambda = \mathrm{64.1 \cdot 10^{-10}\ m}$
$R_\infty = \mathrm{10973731.568508\ m^{−1}} \left(= \frac{R_E}{h\ c} = \frac{\mathrm{2.18\ 10^{-18}\ J}}{h\ c} \right)$

With all this in mind, you should be able to solve the equation to get $Z$.


   $$Z^2 = \frac{n_f^2\ n_i^2}{\lambda\ R_\infty\ (n_f^2 - n_i^2)} \xrightarrow{Z&gt;0} Z \approx 4$$