1. Energy levels of hydrogen-like ions

   For a hydrogen-like ion, the energy of the electron in the $n$th orbit is $E_n = -\frac{13.6 Z^2}{n^2}\;\text{eV}$ where $Z$ is the atomic number.

2. For $\mathrm{He}^+$ ion

   Helium has $Z=2$. So, $E_n = -\frac{13.6\times 2^2}{n^2} = -\frac{54.4}{n^2}\;\text{eV}$

3. First excited state

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

   Therefore, $E_2 = -\frac{54.4}{4} = -13.6\;\text{eV}$

4. Ionization energy from first excited state

   Ionization energy is the energy needed to take the electron from that state to $n=\infty$, where energy is $0$.

   Hence, $\text{Ionization energy} = 0 - (-13.6) = 13.6\;\text{eV}$

5. Option check

   • A: $13.60\,\text{eV}$ ✅
   • B: $6.04\,\text{eV}$ ❌
   • C: $48.36\,\text{eV}$ ❌
   • D: $54.40\,\text{eV}$ ❌

Therefore, the correct answer is A.