What is the atomic radius of an element if it crystallises in a $BCC$ structure with an edge length of unit cell $287 \ pm$ (in $pm$)?

If the radius of an atom of an element which forms a body centered cubic unit cell is $173.2 \ pm$, the volume of the unit cell in $cm^3$ is:

Derive the expression to calculate the density of a unit cell.

$A$ certain element crystallises in a $bcc$ lattice of unit cell edge length $27 \mathring{A}$. If the same element under the same conditions crystallises in the $fcc$ lattice,the edge length of the unit cell in $\mathring{A}$ will be .........
(Round off to the Nearest Integer).
[Assume each lattice point has a single atom]
[Assume $\sqrt{3}=1.73, \sqrt{2}=1.41$]

In a $NaCl$ crystal,the distance between $Na^+$ and $Cl^-$ ions is $2.814 \ \overset{\circ}{A}$ and the density of the crystal is $2.167 \ g/cm^3$. Calculate the value of Avogadro's number $(N_A)$.

Sodium metal crystallizes in a body-centered cubic $(BCC)$ lattice with a unit cell edge length of $4 \ \mathring{A}$. The radius of the sodium atom is $..... \times 10^{-1} \ \mathring{A}$ (Nearest integer).