Answer: $2.16 \times 10^6 \; gm^{-3}$

The density is given by the formula $\large\frac{z\;M}{a^3\;N_A}$

Given $a = 564 \;ppm$.

Each unit cell of NaCl has 4 Na$^{-}$ and 4 Cl$^{-}$ ions $\rightarrow z = 4$

Avagadro's number $N_A = 6.022 \times 10^{23} /\; mol$

The total mass of NaCl $ = 22.99 + 34.34 = 58.5\; g/mol$

$\Rightarrow$ Density $\rho =\large\frac{z\;M}{a^3\;N_A}$$ = \large\frac{4 \times 58.5 }{(564\times10^{-12})^3\;6.022 \times 10^{23}}$$\; gm^{-3} \approx 2.16 \times 10^6 \; gm^{-3}$