A pure metal crystallizes in a body-centered cubic (BCC) lattice...

In a body-centered cubic (BCC) lattice, the atoms touch each other along the body diagonal.

Let the edge length of the unit cell be $a$. The relationship between $a$ and the host atom radius $R$ is given by:
$\sqrt{3}a = 4R \Rightarrow a = \dfrac{4R}{\sqrt{3}}$

The atoms at the corners of the unit cell are separated by distance $a$ along the edge. The space occupied by the two corner atoms along the edge is $2R$.

The empty space available at the center of the edge is:
Empty space $= a - 2R$

Since the interstitial guest atom fits perfectly at the edge center, its diameter $2r$ is equal to this empty space:
$2r = a - 2R$

Substituting $a = \dfrac{4R}{\sqrt{3}}$:
$2r = \dfrac{4R}{\sqrt{3}} - 2R$

Dividing by 2:
$r = \dfrac{2R}{\sqrt{3}} - R = R \left( \dfrac{2 - \sqrt{3}}{\sqrt{3}} \right)$

Therefore, the ratio $\dfrac{r}{R}$ is $\dfrac{2 - \sqrt{3}}{\sqrt{3}}$.

Answer: $\dfrac{2 - \sqrt{3}}{\sqrt{3}}$