Use Bohr’s quantization condition

For an electron in the $n$th Bohr orbit, $2\pi r_n = n\lambda$ where $r_n$ is the radius of the orbit and $\lambda$ is the de Broglie wavelength.

Use the radius of the $n$th Bohr orbit

For hydrogen atom, $r_n = n^2 a_0$ where $a_0$ is the Bohr radius.

Substitute for the 4th orbit

For $n=4$, $r_4 = 4^2 a_0 = 16a_0$

Now from $2\pi r_4 = 4\lambda$ we get $\lambda = \frac{2\pi r_4}{4}$

Calculate $\lambda$

$\lambda = \frac{2\pi (16a_0)}{4} = 8\pi a_0$

Match with the options

$\lambda = 8\pi a_0$

So the correct option is C.