Question

BCC                                                           FCC
$4r=\sqrt{3a}$                                                    $4r\sqrt{2a}$
$a=\frac{4r}{\sqrt{3}}$                                                       $a=\frac{4r}{\sqrt{2}}$
$\frac{d_{BCC}}{d_{FCC}}=\frac{{\displaystyle \frac{Z_{BCC}\times M}{N_A{a}^3}}}{{\displaystyle \frac{Z_{FCC}\times M}{N_A{a}^3}}}=\frac{{\displaystyle \frac{2\times M}{N_A{\left({\displaystyle \frac{4r}{\sqrt{3}}}\right)}^3}}}{{\displaystyle \frac{4\times M}{N_A\times {\left({\displaystyle \frac{4r}{\sqrt{2}}}\right)}^3}}}=\frac{3}{4} \sqrt{\frac{3}{2}}$