If $NaCl$ is doped with ${10}^{-4}mol\%$of $SrC{l}_2$ the concentration of cation vacancies will be $\left(N_A=6.02\times {10}^{23}mo{l}^{-1}\right)$

Doping of $NaCl$ with ${10}^{-4}\text{mol}\%$ of $SrC{l}_2$ means,
100 moles of NaCl are doped with ${10}^{-4}$ mol of $SrC{l}_2$.
$\therefore$ 1 mol of NaCl is doped with ltBrgt $SrC{l}_2=\frac{{10}^{-4}}{100}={10}^{-6}\text{mole}$
As each $S{r}^{2+}$ ion introduces one cation vacancy.
$\therefore$ Concentration of cation vacancies
${10}^{-6}\text{mol/mol of }NaCl$
$={10}^{-6}\times 6.023\times {10}^{23}{\text{mol}}^{-1}$
$=6.023\times {10}^{17}{\text{mol}}^{-1}$