The ratio of the radii of the $2^{nd}$ Bohr orbit of hydrogen atom to $3^{rd}$ Bohr orbits of $L{i}^{2+}$ is:

The correct option is D $4:3$
We know,
$r_n=r_o\times \frac{n^2}{Z}$,

for $2^{nd}$ Bohr orbit of hydrogen atom
Here $Z=1,n=2$
$r_{2\left(H\right)}=r_o\times \frac{2^2}{1}=4r_o$

for $3^{rd}$ Bohr orbit of $L{i}^{2+}$
Here $Z=3,n=3$
$r_{3\left(L{i}^{2+}\right)}=r_o\times \frac{3^2}{3}=3r_o$
So the ratio is,
$\frac{r_{2\left(H\right)}}{r_{3\left(L{i}^{2+}\right)}}=\frac{4r_o}{3r_o}=\frac{4}{3}$