Applying Bohr's model for Helium atom
What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.
Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...
The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).
The experimental value is 1.81 (24.6 eV), which is significantly lower.
Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.