$\mathrm{NaCl}$ of density $2.165\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ is doped with $\mathrm{SrCl}{}_2$ so that $2.0\times 10^{-3}$ mol % of the cations are $\mathrm{Sr}{}^{2+}$ (i.e. $\mathrm{Sr}{}^{2+}$ is a mole fraction of $2.0\times 10^{-5}$ of the cations). Treat the crystal as $\mathrm{NaCl}$ with molar mass $58.5\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $N_A=6.022\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$. Find the number of cation vacancies per $\mathrm{c}{\mathrm{m}}^3$ in units of $10^{17}$ (two decimals).

Numerical answer type, answer in x10^17 per cm^3, tolerance 0.01.