The de Broglie wavelength of an electron in the

$2 \pi r=n \lambda$

$r=\frac{n^{2} a_{0}}{Z}$

$2 \pi \times \frac{4^{2}}{1} a_{0}=4 \lambda$

$\lambda=2 \pi \times \frac{4}{1} a_{0}$

$\lambda=8 \pi a_{0}$