1. Identify the species

   The ion is $\mathrm{Li^{++}}$, which is a hydrogen-like ion with: $Z=3$ because lithium has atomic number $3$ and $\mathrm{Li^{++}}$ has only one electron left.

2. Use the energy formula for a hydrogen-like atom

   For a hydrogen-like species, the energy of the electron in the $n$th orbit is: $E_n=-13.6\frac{Z^2}{n^2}\,\text{eV}$

   Here, for $\mathrm{Li^{++}}$: $E_n=-13.6\frac{3^2}{n^2}=-13.6\frac{9}{n^2}\,\text{eV}$

3. First excited state

   The first excited state corresponds to: $n=2$

   So, $E_2=-13.6\frac{9}{4}=-30.6\,\text{eV}$

4. Energy required to remove the electron

   To remove the electron from this state, we must take it from $E_2=-30.6\,\text{eV}$ to $0\,\text{eV}$.

   Therefore, required energy is: $\Delta E=0-(-30.6)=30.6\,\text{eV}$

5. Check options

   • A: $30.6\,\text{eV}$ ✅
   • B: $13.6\,\text{eV}$ ❌
   • C: $3.4\,\text{eV}$ ❌
   • D: $122.4\,\text{eV}$ ❌

Thus, the correct answer is Option A.