If the radius of first Bohr's orbit is x , then de Broglie...

If ${r}_{3} =$ Radius of ${3}^{rd}$ orbit

and ${r}_{1} =$ Radius of ${1}^{st}$ orbit

${r}_{3} = {r}_{1} \times n{}^{2} = \text{x} \times {3}^{2} = \text{9x}$

$∵ 2 {πr}_{3} = n \lambda$

$∴ \text{\lambda} = \dfrac{2{πr}_{3}}{n } = \dfrac{2\text{\pi} \times 9\text{x}}{3} = 6\text{\pi x}$