The doubly ionized $\mathrm{Li}$ atom is $\mathrm{H}$-like atom

$E_n=-\frac{13.6Z^2}{n^2} \mathrm{eV}$

The ground state energy of the doubly ionised $\mathrm{Li}$ atom

$(Z = 3, n = 1)$ will be

$E_1=\frac{-13.6\times 9}{1}=-122.4 \mathrm{eV}$

Hence, the ionisation energy for an electron in the ground state of doubly ionised $\mathrm{Li}$ atom $=122.4 \mathrm{eV}$