Use Bohr’s quantization condition

For an electron in the $n^{\text{th}}$ orbit, $2\pi r_n = n\lambda$ where $\lambda$ is the de-Broglie wavelength.

Radius of the $n^{\text{th}}$ orbit in hydrogen

For hydrogen, $r_n = n^2 a_0$

Electron in the second orbit

For the second orbit, $n=2$, so $r_2 = 4a_0$

Apply the standing wave condition

Using $2\pi r_n = n\lambda$ for $n=2$, $2\pi r_2 = 2\lambda$ $\lambda = \pi r_2$

Substituting $r_2 = 4a_0$, $\lambda = \pi(4a_0) = 4\pi a_0$

Match with the options

$\lambda = 4\pi a_0$

This corresponds to option D if we take the electron specifically in the second orbit.

About the given options with parameter $n$

The wording says “second orbit”, so the principal quantum number is fixed at $n=2$. In that case, the answer is unambiguously $\lambda = 4\pi a_0$

Option A gives $\frac{8\pi a_0}{n}$ and for $n=2$ this also becomes $4\pi a_0$

So option A matches only after substituting $n=2$. But as written, the direct expression for the second orbit is simply $4\pi a_0$, i.e. option D.

General check

In general, $\lambda_n = \frac{2\pi r_n}{n} = \frac{2\pi (n^2 a_0)}{n} = 2\pi n a_0$

For $n=2$, $\lambda = 2\pi (2)a_0 = 4\pi a_0$

Same result confirmed.