The de-Broglie wavelength ( λ _ B ) associated with the electron...

We know that wavelength is inversely proportional to momentum, let ${\lambda}_{B}$ is de-Broglie Wavelength corresponds to momentum ${P}_{B}$, similarly let ${\lambda}_{G}$ is Wavelength at ground state corresponds to momentum${P}_{G}$,
$\dfrac{{\lambda}_{B}}{{\lambda}_{G}} = \dfrac{{P}_{G}}{{P}_{B}} = \dfrac{m{v}_{G}}{m{v}_{B}}$, and velocity, $V ∝ \dfrac{z}{n}$, where $Z$ is the atomic number, and $n$ is a number of orbits, here, ${n}_{G} = 1$ (ground state), ${n}_{B} = 3$ (second excited state). So, $\dfrac{{\lambda}_{B}}{{\lambda}_{G}} = \dfrac{{n}_{B}}{{n}_{G}} = \dfrac{3}{1}$$⇒ {\lambda}_{B} = 3{\lambda}_{G}$