If Al³+ replaces Na+ at the edge centre of NaCl lattice then the vacancies in 1mole NaCl are: 

In NaCl lattice, ${\mathrm{Cl}}^{-}$ ions forms FCC and ${\mathrm{Na}}^{+}$ ions are present at edge centers and body center.

Number of ${\mathrm{Na}}^{+}$ ions present = $\frac{1}{4}\times 12=3$

Now, charge on ${\mathrm{Al}}^{3+}$= +3 and charge on ${\mathrm{Na}}^{+}$= 1.

So, each ${\mathrm{Al}}^{3+}$ion will replace $3{\mathrm{Na}}^{+}$ions and two vacant sites will be created.

Now, 3/4 moles of ${\mathrm{Na}}^{+}$ ions will be replaced by 1/4 moles of ${\mathrm{Al}}^{3+}$ions to maintain electrical neutrality. Now, remaining 1/4 moles of ${\mathrm{Na}}^{+}$ ions and 1/4 moles of ${\mathrm{Al}}^{3+}$ion will occupy 1/2 moles of lattice sites.

So, Number of vacancies = $\frac{1}{2}\mathrm{moles}=\frac{1}{2}\times 6.023\times {10}^{23}=3.0115\times {10}^{23}$