Given density of  NaCl is $2.17\times {10}^3{\mathrm{kgm}}^{-3}$. Thus edge length of   crystal unit cell is

  NaCl has fcc structure Thus, number of   molecules in a unit cell = 4
$\mathrm{Density} =\frac{ \mathrm{Mass} \mathrm{of} 4\left(\mathrm{NaCl}\right) \mathrm{molecules} }{ \mathrm{Volume} \mathrm{of} \mathrm{cube} } \mathrm{Molar} \mathrm{mass} \mathrm{of}$
$\begin{array}{l}\mathrm{NaCl}=58.5{\mathrm{gmol}}^{-1}=58.5\times {10}^{-3}{\mathrm{kgmol}}^{-1}\mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{molecule} =\frac{58.5\mathrm{x}{10}^{-3}}{6.02 \mathrm{x}{10}^{23}}\mathrm{kg}\\ \therefore 2.17\times {10}^3{\mathrm{kgm}}^{-3}=\frac{4\times 58.5\times {10}^{-3}}{6.02\times {10}^{23}\times {\mathrm{a}}^3}\therefore {\mathrm{a}}^3=\frac{4\times 58.5\times {10}^{-3}}{6.02\times {10}^{23}\times {10}^3}\\ =1.791\times {10}^{-28}{\mathrm{m}}^3\mathrm{a}=5.64\times {10}^{-10}\mathrm{m}=564\times {10}^{-12}\mathrm{m}=564\mathrm{pm}\end{array}$