Let the first ionization energy of helium be $E_1$.

This is the energy required to remove one electron from a neutral helium atom: $\text{He} \to \text{He}^+ + e^-$

Given relation for singly ionized helium

The energy required to remove the electron from $\text{He}^+$ is $2.2$ times the energy required to remove an electron from He atom.

So, if the second ionization energy is $E_2$, then $E_2 = 2.2E_1$

Use hydrogen-like atom formula for $\text{He}^+$

A singly ionized helium atom $\text{He}^+$ is a hydrogen-like ion with atomic number $Z=2$.

Its ground state ionization energy is $E_2 = 13.6 Z^2 = 13.6 \times 2^2 = 54.4\,\text{eV}$

Find the first ionization energy

From $54.4 = 2.2E_1$ we get $E_1 = \frac{54.4}{2.2}$ $E_1 \approx 24.73\,\text{eV}$

Total energy to ionize helium completely

This is the sum of first and second ionization energies: $E_{\text{total}} = E_1 + E_2$ $E_{\text{total}} = 24.73 + 54.4 = 79.13\,\text{eV}$

Match with options

$E_{\text{total}} \approx 79\,\text{eV}$

Therefore, the correct option is C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

Hence, they agree.