Energy required to remove an electron from He
We can use Bohr's atomic model to calculate the ionization energy of $\ce{He+}$.
The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives $IE_2=E_{\infty}-E_1$ where $E_1$ is the energy in the first bohr-orbit (principal quantum number=$1$). Conventionally taking $E_{\infty}$ as $0$ we use the Bohr's equation to find the energy $E_1$:-
$$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$
 where $Z$= atomic number, $e$= charge on electron, $n$= principal quantum number, $h$= Planck's constant, $m$= mass of electron, $\epsilon_0$= permittivity of free space.   

Using $n=1$, $Z=2$ and the appropriate values for other constants we get $IE_2=-E_n=\pu{54.4eV}$.
Now the energy for removing both electron $IE_{net}=IE_1+IE_2$. Using $IE_{net}=\pu{79eV}$and $IE_2=\pu{54.4eV}$ we get, $$IE_1=\pu{24.6eV}$$
which corresponds to option (A), which I assume is typed wrong by you in the question.