\(r=r_{0} \frac{n^{2}}{Z}\)

For the radius to be equal to radius of hydrogen atoms, \(r=r_{0}\)

Therefore,

\(r_{0}=r_{0} \frac{n^{2}}{Z}\)

\(n=\sqrt{Z}\)

\(=\sqrt{4}\)

\(=2\)

First excited state.

Thus, at \(first\) excited state of \(Be ^{3+}\) radius of \(e ^{-}\) will be same as \(H\) atoms and electron in ground state.