Ratio of de-Broglie wavelength to orbital radius of an electron $L{i}^{2+}$ in first excited state is  $\frac{6\pi }{x}.$ Find x.

de-Broglie wavelength of electron in Bohr’s atom is given by
$\lambda =\frac{2\pi r}{n}\Rightarrow \frac{\lambda }{r}=\frac{2\pi }{n}$
 For 1st excited state, n = 2
$\Rightarrow \frac{\lambda }{r}=\pi =\frac{6\pi }{6}\Rightarrow x=6$