Answer
Attractive electric force between the ions is $2.8972\times10^{-9}N$
Work Step by Step
Singly charge sodium means one electron is removed from the sodium atom
so charge on sodium ion is $q_{Na^+}= +e=+1.6\times10^{-19}C$
similarly singly charged chlorine ions means one electron is given to chlorine atom
so charge on chlorine ion $q_{Cl^-}= -e=-1.6\times10^{-19}C$
Distance between sodium and chlorine ions in crystal is given as
$r= 2.82\times10^{-10}m$
so from coulombs law magnitude of attractive force between the ions will be $F=k\frac{q_{Na^+}\times q_{Cl^-}}{r^2}$
putting magnitudes of $q_{Cl^-}$, $q_{Na^+}$ as as $1.6\times10^{-19}C$ , $k=9.0\times10^{9}N.m^2/C^2$, $r= 2.82\times10^{-10}m$
$F=9.0\times10^{9}N.m^2/C^2\times \frac{1.6\times10^{-19}C\times 1.6\times10^{-19}C}{(2.82\times10^{-10}m)^2}$ =$2.8972\times10^{-9}N$