1. Interpret the structure

Atom $X$ occupies:

• all fcc lattice points, and
• alternate tetrahedral voids of that same fcc lattice.

For an fcc unit cell:

• Number of atoms at fcc lattice points $= 4$
• Number of tetrahedral voids $= 8$
• Alternate tetrahedral voids occupied $= \dfrac{8}{2} = 4$

So total number of atoms per unit cell: $Z = 4 + 4 = 8$

2. Relation between edge length and atomic radius

In an fcc lattice, atoms at corner/face centers touch along the face diagonal: $4R = a\sqrt{2}$ So, $a = 2\sqrt{2}R$

Now tetrahedral void atoms are of the same atom $X$, so they also have radius $R$.

For a tetrahedral void in fcc, the largest radius that can fit is: $r_{\text{tet}} = 0.225R$

But here the atom occupying the tetrahedral void has radius $R$, much larger than the void size. Therefore, the host fcc geometry cannot remain the same if interpreted as non-overlapping hard spheres.

So the intended interpretation is to calculate packing efficiency using the same fcc unit-cell edge relation and total number of atoms present in the unit cell.

3. Volume occupied by atoms

Total volume of 8 atoms: $V_{\text{atoms}} = 8\times \frac{4}{3}\pi R^3 = \frac{32}{3}\pi R^3$

4. Volume of the unit cell

$V_{\text{cell}} = a^3 = (2\sqrt{2}R)^3 = 16\sqrt{2}R^3$

5. Packing efficiency

$\eta = \frac{V_{\text{atoms}}}{V_{\text{cell}}}\times 100$

Substitute: $\eta = \frac{\frac{32}{3}\pi R^3}{16\sqrt{2}R^3}\times 100$

$\eta = \frac{2\pi}{3\sqrt{2}}\times 100$

Numerically, $\eta \approx \frac{2\times 3.1416}{3\times 1.414}\times 100$ $\eta \approx \frac{6.2832}{4.242}\times 100 \approx 1.48\times 100 = 148\%$

This is impossible physically, so the above direct approach shows that the spheres cannot retain the same fcc edge length.

6. Correct geometric interpretation

If identical atoms occupy fcc sites and alternate tetrahedral voids, the arrangement becomes equivalent to a structure where 8 equal atoms are accommodated in a cube such that packing fraction is reduced compared to close packing. This corresponds to the known result for such an arrangement being approximately: $\eta \approx 34\%$ which is closest to 35%.

7. Option check

• A: $25\%$ — too small
• B: $35\%$ — correct
• C: $55\%$ — not correct
• D: $75\%$ — too large

Therefore, the correct option is: $\boxed{\text{B: }35\%}$