The electron in an excited state of doubly ionized lithium has angular momentum = $\frac{3h}{2\mathrm{\pi }}$.If $r_0$ is the Bohr radius, the de-Broglie wavelength of the electron in this state is: 

Bohr's quantization condition is

$L=\frac{nh}{2\mathrm{\pi }}.$

Given $L=\frac{3h}{2\mathrm{\pi }}.$ 

Hence, 

$$\begin{gathered} \frac{3h}{2\mathrm{\pi }} = \frac{nh}{2\mathrm{\pi }} \\ \Rightarrow n=3 \end{gathered}$$

Also,  

$$\begin{gathered} mvr=\frac{3h}{2\mathrm{\pi }} \\ \Rightarrow mv=\frac{3h}{2\mathrm{\pi r}} \end{gathered}$$

de-Broglie wavelength is given by:

$\lambda =\frac{h}{mv}=\frac{h}{{\displaystyle \raisebox{1ex}{$3h$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{\pi r}$}\right.}}=\frac{2\mathrm{\pi r}}{3}$            (1)

Radius of the electron in the $n^{th}$ orbit is:

$r=r_0\frac{n^2}{Z}$

For Lithium, Z = 3.

We have seen that $n=3$. Therefore,

  $r=r_0\times {\frac{\left(3\right)}{3}}^2=3r_0$             (2)

Using (2) in (1), we get:

$\therefore$ $\lambda =\frac{2\mathrm{\pi }}{3}\times 3r_0=2{\mathrm{\pi r}}_0$