If $\text{NaCl}$ is doped with $10^{-4}\text{ mol\%}$ of ${\text{SrCl}}_2$, the concentration of cation vacancies will be ($N_A=6.02\times 10^{23}{\text{ mol}}^{-1}$):

In $\text{NaCl}$, each ${\text{Sr}}^{2+}$ ion replaces two ${\text{Na}}^{+}$ ions to maintain electrical neutrality. One site is occupied by ${\text{Sr}}^{2+}$ and the other remains vacant. Thus, 1 mole of ${\text{SrCl}}_2$ creates 1 mole of cation vacancies.Given doping $=10^{-4}\text{ mol\%}=\frac{10^{-4}}{100}=10^{-6}\text{ moles}$ of ${\text{SrCl}}_2$ per mole of $\text{NaCl}$.Number of vacancies $=\text{moles}\times N_A=10^{-6}\times 6.02\times 10^{23}=6.02\times 10^{17}$.