To solve this problem, we need to find the ratio of the density of iron in bcc structure to that in fcc structure.$\\$The density $\rho$ of a crystal is given by:$\\ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \\$where:$\\$• $Z$ is the number of atoms per unit cell.$\\$• $M$ is the molar mass.$\\$• $a$ is the edge length of the unit cell.$\\$• $N_A$ is Avogadro's number.$\\$For bcc structure:$\\ Z_{bcc} = 2 \\$For fcc structure:$\\ Z_{fcc} = 4 \\$The relationship between the edge length $a$ and the atomic radius $r$ is:$\\$• For bcc: $a_{bcc} = \frac{4r}{\sqrt{3}} \\$• For fcc: $a_{fcc} = \frac{2\sqrt{2}r}{1} \\$Now, let's calculate the density ratio:$\\$Density of bcc: $\rho_{bcc} = \frac{Z_{bcc} \cdot M}{a_{bcc}^3 \cdot N_A} \\$Density of fcc: $\rho_{fcc} = \frac{Z_{fcc} \cdot M}{a_{fcc}^3 \cdot N_A} \\$The ratio of densities is:$\\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{\frac{Z_{bcc} \cdot M}{a_{bcc}^3 \cdot N_A}}{\frac{Z_{fcc} \cdot M}{a_{fcc}^3 \cdot N_A}} \\$Simplifying, we get:$\\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{Z_{bcc} \cdot a_{fcc}^3}{Z_{fcc} \cdot a_{bcc}^3} \\$Substitute the values:$\\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{2 \cdot \left(\frac{2\sqrt{2}r}{1}\right)^3}{4 \cdot \left(\frac{4r}{\sqrt{3}}\right)^3} \\$Simplify the expression:$\\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{2 \cdot \left(8\sqrt{2}r^3\right)}{4 \cdot \left(\frac{64r^3}{3\sqrt{3}}\right)} \\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{16\sqrt{2}r^3}{\frac{256r^3}{3\sqrt{3}}} \\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{16\sqrt{2} \cdot 3\sqrt{3}}{256} \\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{48\sqrt{6}}{256} \\ \frac{\rho_{bcc}}{\rho_{fcc}} = \frac{3\sqrt{3}}{4\sqrt{2}} \\$Therefore, the correct option is Option 1: $\frac{3\sqrt{3}}{4\sqrt{2}}.$