Explanation

(c) The energy of $n$th orbit of hydrogen like atom is,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{n}\ = \ - 13.6\ \frac{Z^{2}}{n^{2}}$
Here,$\ Z = 11$ for Na atom. 10 electrons are removed already. For the last electron to be removed $n$=1.
$\therefore E_{n} = \frac{- 13.6\ \times {(11)}^{2}}{{(1)}^{2}}\ eV$
= $- 13.6\ \times {(11)}^{2}$ eV