At which excited state of $Be^{3+}$ will the radius of the $e^{-}$ be the same as that of an $H$ atom in the ground state?

Four electrons,each of mass $m_{e}$,are in a one-dimensional box of size $L$. Assume that the electrons are non-interacting,obey the Pauli exclusion principle,and are described by standing de Broglie waves confined within the box. Define $\alpha = h^{2} / 8 m_{e} L^{2}$ and $U_{0}$ to be the ground state energy. Then,

If an electron in a hydrogen atom jumps from the third orbit to the second orbit,the frequency of the emitted radiation is given by (where $c$ is the speed of light and $R$ is the Rydberg constant):

The angular momentum of an electron in the $n^{th}$ orbit is given by:

According to Bohr's theory,the angular momentum of an electron revolving in the second orbit of a hydrogen atom will be:

In the Bohr model of a hydrogen-like atom,the force between the nucleus and the electron is modified as $F = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right)$,where $\beta$ is a constant. For this atom,the radius of the $n^{th}$ orbit in terms of the Bohr radius $\left( a_0 = \frac{\varepsilon_0 h^2}{m \pi e^2} \right)$ is: