The ratio of radii of second orbits of  $L{i}^{2+}$ and  $B{e}^{3+}$ is 

The radius of an electron (r_n​) in any orbit can be calculated as follows 

${\mathrm{r}}_{\mathrm{n}}=\frac{0.53}{\mathrm{z}}{\mathrm{xn}}^2$
Where, n= number of orbits,
Z= atomic number
For Li²⁺−----n=2,Z=3
For Be³⁺−-----n=2,Z=4
Therefore,

$$\begin{gathered} \frac{\mathrm{r}\left({\mathrm{Li}}^{2+}\right)}{ \mathrm{r}\left({\mathrm{Be}}^{3+}\right)}=\left[\frac{{\mathrm{n}}^2}{\mathrm{Z}}\right]\left({\mathrm{Li}}^{2+}\right)\times [\frac{\mathrm{Z}}{{\mathrm{n}}^2}]\left({\mathrm{Be}}^{3+}\right) \\ \frac{\mathrm{r}\left({\mathrm{Li}}^{2+}\right)}{ \mathrm{r}\left({\mathrm{Be}}^{3+}\right)}= [ \frac{2^2}{ 3} ]\times [ \frac{4}{2^2} ] \\ \frac{\mathrm{r}\left({\mathrm{Li}}^{2+}\right)}{ \mathrm{r}\left({\mathrm{Be}}^{3+}\right)}=\frac{4}{3} \end{gathered}$$