Radius of $2^{\mathrm{nd}}$ orbit of ${\mathrm{B}}^{4+}$ ion is given as,

$\mathrm{r}=0.529 \frac{{\mathrm{n}}^2}{\mathrm{Z}}$ $\stackrel{°}{\mathrm{A}}$

Where, $\mathrm{n}$ = Number of shells and $\mathrm{Z}=$Atomic number

This formula is only applicable for hydrogen and hydrogen like species i.e., species containing a single electron.

For ${\mathrm{B}}^{4+}$, $\mathrm{n}=2, \mathrm{Z}=5$

$\mathrm{r}=0.529\frac{2^2}{5}=0.529\times \frac{4}{5}=0.4232 \stackrel{°}{\mathrm{A}}$