If the radius of first Bohr orbit is x, then de Broglie wavelength of electron in 3rd orbit is nearly.

$\begin{array}{l}{\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{a}}_0}{\mathrm{Z}},\therefore {\mathrm{r}}_0\propto {\mathrm{n}}_2\\ \frac{{\mathrm{r}}_3}{{\mathrm{r}}_1}=\frac{9}{1}\\ {\mathrm{r}}_3=9{\mathrm{r}}_1=9\mathrm{x}\\ \mathrm{Also}{\mathrm{mvr}}_{\mathrm{n}}=\frac{\mathrm{nh}}{2\mathrm{\pi }}\\ {\mathrm{mvr}}_3=\frac{3\mathrm{h}}{2\mathrm{\pi }}\\ \mathrm{mv}=\frac{\mathrm{h}}{6\mathrm{\pi x}}\\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}=6\mathrm{\pi x}\end{array}$