The percentage of void space of a metallic element crystallising in a ABCABC ..... type lattice pattern is

The ABCABC type of arrangement leads to a face-centered cubic structure that is similar to a cubic close packing in general. Atoms can be found on the cube's corners and face centres in FCC.
Consequently, a cubic unit cell has a total of, $\mathrm{n}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

Atoms on face diagonals will now come into contact with one another in FCC. So, $4r=\sqrt{2a}$
$$\begin{gathered} r=\frac{a}{2\sqrt{2}}\left(r\text{ is radius of each atom }\right) \\ \text{Currently, cubic unit cell volume by edge length }a=a^3 \\ Vol. occupied = 4\times \frac{3}{4}\pi {\left(r\right)}^3 \\ \begin{array}{l}=4\times \frac{4}{3}\times \pi {\left(\frac{\mathrm{a}}{2\sqrt{2}}\right)}^3\\ =\frac{16}{3}\times \pi \times {\mathrm{a}}^{3}16\sqrt{2}\\ =\frac{\pi {\mathrm{a}}^3}{3\sqrt{2}}\\ \text{→ Volume occupied }=\frac{\frac{\pi {\mathrm{a}}^3}{{\mathrm{a}}^3}}{{\mathrm{a}}^3}\times 100=74\%\\ \text{→ Void }=(100-74)\%=26\%\end{array} \end{gathered}$$
 

Therefore, the correct option is B 26%