A He$$^+$$ ion is in its first excited state. Its ionization energy is:

For a hydrogen-like ion, such as $$\text{He}^+$$ whose nucleus has charge number $$Z=2$$, the energy of an electron in the $$n^{\text{th}}$$ Bohr orbit is given by the well-known formula

$$E_n=-\,13.6\,\text{eV}\;\frac{Z^{2}}{n^{2}}.$$

We have an electron in the first excited state, which corresponds to the principal quantum number $$n=2$$ (because the ground state is $$n=1$$ and the next level is $$n=2$$).

Substituting $$Z=2$$ and $$n=2$$ into the formula, we get

$$E_2=-\,13.6\,\text{eV}\;\frac{(2)^{2}}{(2)^{2}}=-\,13.6\,\text{eV}.$$

The negative sign simply indicates that the electron is bound. To remove (ionize) the electron completely, we must supply energy equal in magnitude to this binding energy while taking the electron to $$E=\;0$$ at infinity. Hence, the ionization energy from this state is

$$\text{Ionization energy}=|E_2|=13.6\,\text{eV}.$$

Among the given options, 13.60 eV is listed as option A.

Hence, the correct answer is Option A.