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$d = \frac{ZM}{a^3N_A}$ $\text{Consider the bcc unit cell. The density of iron in bcc } d_1 = \frac{Z_1M}{a_1^3N_A}$ $\text{Substituting the values, we get } d_1 = \frac{2M}{\left(\frac{4}{\sqrt{3}}r\right)^3}N_A$ $\text{Consider the fcc unit cell. The density of iron in fcc } d_2 = \frac{Z_2M}{a_2^3N_A}$ $\text{Substituting the values, we get } d_2 = \frac{4M}{\left(\frac{4}{\sqrt{2}}r\right)^3}N_A$ $\text{Now we have to calculate the ratio of iron at room temperature and at } 900 \, ^\circ \text{C}$ $\frac{d_1}{d_2} = \frac{\frac{2M}{\left(\frac{4}{\sqrt{3}}r\right)^3}N_A}{\frac{4M}{\left(\frac{4}{\sqrt{2}}r\right)^3}N_A}$ $\text{Taking reciprocal, we get}$ $\frac{d_1}{d_2} = \frac{2M}{4M} \times \frac{\left(\frac{4}{\sqrt{2}}r\right)^3}{\left(\frac{4}{\sqrt{3}}r\right)^3}$ $\text{Canceling the common terms we get}$ $\frac{d_1}{d_2} = \frac{\left(\frac{4}{\sqrt{2}}r\right)^3}{\left(\frac{4}{\sqrt{3}}r\right)^3} \times \frac{2}{4} = \left(\frac{4}{\sqrt{2}}\right)^3 \times \left(\frac{\sqrt{3}}{4}\right)^3 \times \frac{1}{2} = \frac{\sqrt{3}}{\sqrt{3}} \times \frac{1}{2}$ $\text{On simplification,}$ $\frac{d_1}{d_2} = \frac{3\sqrt{3}}{2\sqrt{2}} \times \frac{1}{2} = \frac{3\sqrt{3}}{4\sqrt{2}}$ $\text{Hence the ratio of density is } \frac{3\sqrt{3}}{4\sqrt{2}}$