The binding energy of an electron in the ground state of He is equal to 24.6 eV. The binding energy required to remove both the electrons is

Energy required to remove 1st e^- from He
${\mathrm{E}}_1=+13.6\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}=+13.6\times \frac{4}{1}(\mathrm{Z}=2)=54.5\mathrm{eV}$
Energy required to remove 2nd e^– from He E₂ = 24.6eV
The binding energy required to remove both the electrons is
E₁ + E₂ = 54.4 + 24.6eV = 79eV