The energy of the $n^{\text{th}}$ state of Hydrogen atom $= E_n = -\frac{13.6}{n^2}$ For ground state, $n=1$ When atom is in second excited state, $n=3$ $\therefore E = -\frac{13.6}{3^2} - 1.51\,\text{eV}$ Now, the de Broglie wavelength associated with an electron is $\lambda = \frac{h}{p}$ $h=$ Planck's constant $p=$ Momentum of electron $m=$ Mass of electron $p = \sqrt{2 m E}$ $\therefore \lambda = \frac{h}{p}$ $\text{Or } \lambda = \frac{h}{\sqrt{2 m E}}$ $\therefore \lambda = \frac{6.6 \times 10^{-34} \times 10^{25}}{6.631}$$= 0.995 \times 10^9\,\text{m}$$= 9.95\,\text{A}^{\circ}$