For hydrogen and hydrogen like atoms
                  ${E }_n=-13\text{.}6\frac{\left({\text{Z}}^2\right)}{\left(n^2\right)}\text{eV}$ 
Therefore,ground state energy of doubly ionized lithium atom (Z=3,n=1) will be

$E_n=-13·6\frac{{\left(3\right)}^2}{\frac{{\left(3\right)}^2}{{\left(1\right)}^2}}=-122·\mathrm{4 }\text{e}\text{V}$
$\therefore$ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV