Energy required for the electron excitation in $$\text{Li}^{++}$$ from the first to the third Bohr orbit is:

For any hydrogen-like ion the Bohr energy of the $$n^{\text{th}}$$ orbit is given by
$$E_n = -13.6\,\text{eV}\,\frac{Z^{2}}{n^{2}}$$ where $$Z$$ is the atomic number.

For $$\text{Li}^{++}$$, $$Z = 3$$ because two electrons have been removed but the nucleus still contains three protons.

Energy of the 1st orbit (ground state):
$$E_1 = -13.6 \times \frac{3^{2}}{1^{2}} = -13.6 \times 9 = -122.4 \,\text{eV}$$

Energy of the 3rd orbit:
$$E_3 = -13.6 \times \frac{3^{2}}{3^{2}} = -13.6 \times \frac{9}{9} = -13.6 \,\text{eV}$$

The excitation energy required to lift the electron from $$n = 1$$ to $$n = 3$$ is the difference between these two levels (final minus initial):
$$\Delta E = E_3 - E_1 = (-13.6) - (-122.4) = 108.8 \,\text{eV}$$

Thus the energy needed is $$108.8 \,\text{eV}$$.

Option B which is: $$108.8 \, \text{eV}$$