Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

For BCC lattice : $Z = 2$, $a = \frac{4r}{\sqrt{3}}$

For FCC lattice : $Z = 4$, $a = 2\sqrt{2} r$

$\therefore \frac{d_{25^\circ\text{C}}}{d_{900^\circ\text{C}}} = \frac{\left(\frac{ZM}{N_A a^3}\right)_{\text{BCC}}}{\left(\frac{ZM}{N_A a^3}\right)_{\text{FCC}}}$

$= \frac{2}{4} \left( \frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}} \right)^3$

$= \frac{3\sqrt{3}}{4\sqrt{2}}$