The molarity of solution with respect to $O{H}^{-}$ assuming no loss due to evaporation is:

Also Eq. of $O{H}^{-}$ formed = Eq. of $C{l}_2$ formed
            = $10^3$ /35.5=28.17
$\therefore$      Mole of   $O{H}^{-}$ formed =28.17       ( $\because$ mole = Eq./valence factor)

$\therefore$                             
 [ $O{H}^{-}$ ] = $\frac{Mole}{\text{Volume in litre }}$ = $\frac{28.17}{20}$ =$1.408\text{ mol }litr{e}^{-1}$
Note: 20% by mass solution of NaOH contains nearly 4.27 mole of NaCl per litre, i.e., sufficient NaCl is there.