$$\begin{gathered} \mathrm{Concentratoin} \mathrm{of} {\mathrm{Al}}^{+3} \mathrm{ions}=\frac{{10}^{-3}}{100}={10}^{-5} \\ 1 \mathrm{mole} =6.023\times {10}^{23}{\mathrm{Al}}^{+3} \mathrm{ions} \\ {10}^{-5 }\mathrm{mole}=.......? \\ ={10}^{-5}\times 6.023\times {10}^{23}=6.023\times {10}^{18} \end{gathered}$$

One aluminium ion will remove 3 sodium ions to maintain electrical neutrality,

hence number of sodium ions removed from crystal $=3\times 6.023\times {10}^{18}$