To find the atomic radius of an element with a body-centered cubic (bcc) structure, we need to understand the geometry of the bcc unit cell.$\\$In a bcc structure, the atoms are located at the corners of the cube and one atom is at the center. The relation between the atomic radius $r$ and the edge length $a$ of the cube is given by:$\\$• The body diagonal of the cube contains $2$ atomic radii plus the diameter of the central atom.$\\$• The body diagonal of the cube is $\sqrt{3}a. \\$• The body diagonal also equals $4r$ (since it passes through the center of the atom at the center of the cube).$\\$Thus, we have the equation:$\\ \sqrt{3}a = 4r \\$Solving for $r: \\ r = \frac{\sqrt{3}a}{4} \\$Given that the cell edge $a = 288$ pm, substitute this into the equation:$\\ r = \frac{\sqrt{3} \times 288}{4}$ pm$\\$Therefore, the atomic radius is given by Option 4: $\frac{\sqrt{3} \times 288}{4}$ pm.