The three distances $\text{AB,AC}$ and $\text{A}{\text{A}}^{\prime }$ in the given BCC lattice are :

Edge length $=\text{A}\text{B}=\text{A}\text{D}=\text{B}\text{C}=\text{C}\text{D}=\text{a}$

$\therefore AC=\sqrt{{\left(\text{AB}\right)}^2+{\left(\text{BC}\right)}^2}=\sqrt{2}a$

Body diagonal is $\mathrm{AG}$.

Consider a triangle $∆\mathrm{ACG}$

$\text{AG}=\sqrt{{\left(\text{AC}\right)}^2 + {\left(\text{CG}\right)}^2}$

$=\sqrt{{\left(\sqrt{2\text{a}}\right)}^2 +{\left(\text{a}\right)}^2}\text{}$

$=\sqrt{3}\text{a}$

Now, ${\text{AA}}^{\prime }+{\text{A}}^{\prime }\text{G}=2{\text{AA}}^{\prime }=\sqrt{3}\text{a}$

$\therefore {\text{AA}}^{\prime }=\frac{\sqrt{3}\text{a}}{2}$