The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

We first recall the hydrogen-like energy formula. For any one-electron atom or ion, the energy of the electron in the orbit having principal quantum number $$n$$ is given by the Bohr formula

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

where $$Z$$ is the atomic number of the nucleus. The ionisation energy (the energy needed to remove the electron completely) is the magnitude of the ground-state energy, i.e. the energy corresponding to $$n = 1$$.

Now, a singly ionised Helium atom is $$\text{He}^+$$. It has only one electron and a nucleus with charge $$Z = 2$$. Setting $$n = 1$$ and $$Z = 2$$ in the above formula, we get the energy that must be supplied to remove this last electron:

$$E_2 \;=\; \left|E_{n=1}\right| \;=\; 13.6 \times \dfrac{(2)^2}{(1)^2}\ \text{eV} \;=\; 13.6 \times 4\ \text{eV} \;=\; 54.4\ \text{eV}$$

This quantity, $$E_2 = 54.4\ \text{eV}$$, is the second ionisation energy of Helium because it removes the second (and last) electron from $$\text{He}^+$$ and produces $$\text{He}^{2+}$$.

The question tells us that this second ionisation energy is 2.2 times the energy required to remove the first electron from a neutral Helium atom. Let the first ionisation energy be $$E_1$$. Then we have

$$E_2 \;=\; 2.2\,E_1$$

Substituting the numerical value of $$E_2$$ just obtained,

$$54.4\ \text{eV} \;=\; 2.2\,E_1$$

Now we solve for $$E_1$$ step by step:

$$E_1 \;=\; \dfrac{54.4\ \text{eV}}{2.2}$$

$$E_1 \;=\; 24.727\ \text{eV}\;(\text{approximately})$$

The total energy required to completely ionise a neutral Helium atom is simply the sum of the first and second ionisation energies, because we must supply $$E_1$$ to remove the first electron and then $$E_2$$ to remove the second:

$$E_{\text{total}} \;=\; E_1 + E_2 \;=\; 24.727\ \text{eV} + 54.4\ \text{eV}$$

$$E_{\text{total}} \;=\; 79.127\ \text{eV} \; \approx \; 79\ \text{eV}$$

Hence, the correct answer is Option B.