The compound M₂O (molar mass= 200 g mol^-1 ) has anti-fluorite closest unit cell. If the radius of M+ is 79.55 pm, the density of M₂Owould be

r_a = r_c/0.225 = (79.55 pm)/0.225 =353.56 pm;

$a =2\sqrt{2} r_{a }=2\sqrt{2} \left( 353.56 pm \right) =1000 pm$

There are 4 molecules of M₂X per unit cell of anti-fluorite structure. Hence 

$\rho =\frac{NM}{N_{\mathrm{A}}{a}^3}=\frac{4\left(200\mathrm{g}{\mathrm{mol}}^{-1}\right)}{\left(6.022\times {10}^{23}{\mathrm{mol}}^{-1}\right){\left(1000\times {10}^{-10}\mathrm{cm}\right)}^3}=1.33\mathrm{g}{\mathrm{mol}}^{-1}$