Electrons shown in the two figures are in same quantum state. Magnitude of electron’s energy in the box is  $\frac{2h^2}{m{l}^2}$. Here, h is planck’s constant,  m is mass of electron,  $l$ is length of the box.
 
Considering that magnitude of electron’s energy in 1st orbit of hydrogen atom is  $E_0$, potential energy of electron in the Bohr’s atom shown is  $\frac{E_0}{x}$. Find  $x$.

If electronic standing wave has n loops, length of the boc can be written as
  $l=\frac{n\lambda }{2}\Rightarrow \lambda =\frac{2l}{n}$
Momentum as per de-Broglie hypothesis,
          $p=\frac{h}{\lambda }=\frac{h}{\frac{2l}{n}}=\frac{nh}{2l}$
Therefore, kinetic energy will be
$$\begin{gathered} K=\frac{p^2}{2m}=\frac{\left(\frac{nh}{2l}\right)}{2m}=\frac{n^2{h}^2}{8m{l}^2} \\ \Rightarrow \frac{2h^2}{m{l}^2}=\frac{n^2{h}^2}{8m{l}^2}\Rightarrow n=4\mathrm{.......}\left(i\right) \end{gathered}$$
Now, potential energy of electron in nth orbit in Bohr’s atom is given by $E=-\frac{2E_0{Z}^2}{n^2}$ 
For  $H{e}^{+},Z=2$ and  $n=4$  [using Eq. (i)]
So, potential energy of electron will be $E=\frac{-2E_0{\left(2\right)}^2}{4^2}=\frac{-E_0}{2}$
$\Rightarrow \left|E\right|=\frac{E_0}{2}\therefore x=2$