$2 \,M$ solution of $NaOH$ means $2$ mole $NaOH$ is present in $1\, L$ solution;
density $= 1.28 \,g/ml$
mass of solution = volume of solution $\times$ density
$= 1200 \times 1.28$
$= 1280\, g$
mass of solvent = mass of solution - mass of solute
$= 1280 - 80$
$= 1200 \,g$
molality $ = \frac{2}{1200} \times 1200$
$ = \frac{20}{12} = \frac{10}{6}$
$ = \frac{5}{3} = 1.67\,m$