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

According to Bohr theory

$\begin{array}{l}\mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}\\ \therefore \frac{\mathrm{h}}{\mathrm{mv}}=\frac{2\mathrm{\pi r}}{\mathrm{n}}\\ \mathrm{According}\mathrm{to}\mathrm{de}-\mathrm{Broglie}\\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}\\ \therefore \mathrm{\lambda }=\frac{2\mathrm{\pi r}}{\mathrm{n}}\end{array}$

where r is the radius of the orbit having principal quantum number = n

$\begin{array}{l}\because \frac{{\mathrm{a}}_3}{{\mathrm{a}}_1}=\frac{9}{1}\left(\mathrm{r}\propto {\mathrm{n}}_2\right)\left(\mathrm{here}{\mathrm{a}}_3=\mathrm{radius}\mathrm{of}3\mathrm{rd}\mathrm{orbit}\right)\\ \therefore \mathrm{\lambda }=\frac{2\mathrm{\pi }\times 9{\mathrm{a}}_1}{3}=6{\mathrm{\pi a}}_1\end{array}$