The ground state energy of a hydrogen atom is -13.6 eV . The energy...

For a hydrogen-like species, the energy of an electron in the $n^{\mathrm{th}}$ orbit is given by $E_n = -13.6 \dfrac{Z^2}{n^2} \mathrm{eV}$.

For a $\mathrm{Li}^{2+}$ ion, the atomic number $Z = 3$.

The first excited state corresponds to the principal quantum number $n = 2$.

Substituting these values, the energy of the electron in the first excited state is:
$E_2 = -13.6 \times \dfrac{3^2}{2^2} = -13.6 \times \dfrac{9}{4} = -30.6 \mathrm{eV}$.

The ionization energy is the energy required to remove the electron from this state to infinity ($E_{\infty} = 0$).
Ionization energy $= E_{\infty} - E_2 = 0 - (-30.6 \mathrm{eV}) = 30.6 \mathrm{eV}$.

Answer: $30.6 \mathrm{eV}$