According to the Bohr model, the first radius for hydrogen atom:

${\mathrm{r}}_1=0.529\mathrm{A}°=\mathrm{x}$

The third radius of the Bohr model:${\mathrm{r}}_3=0.529 {\mathrm{n}}^2 \mathrm{A}° = 0.529{\left(3\right)}^2\mathrm{A}°=9\mathrm{x}$

Hence, the de Broglie wavelength:

$\mathrm{\lambda }=\frac{2{\mathrm{\pi r}}_3}{\mathrm{n}}=\frac{2\times \mathrm{\pi }\times \left(9\mathrm{x}\right)}{3}=6\mathrm{\pi x}$