The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}$ eV . The ground state corresponds to $n=1$. The 'second excited state' corresponds to the principal quantum number $n=3$ (as $n=2$ is the first excited state).

Calculating the energy of the electron in the $n=3$ orbit: $E_3 = -\frac{13.6}{3^2} \text{ eV} = -\frac{13.6}{9} \text{ eV} \approx -1.51 \text{ eV}$.

The ionization energy is the energy required to remove the electron from this state to infinity (where energy is 0). Ionization Energy = $E_{\infty} - E_3 = 0 - (-1.51) \text{ eV} = +1.51 \text{ eV}$.