 Solution

The fluorite structure consists of a face-centered cubic lattice of calcium sites with all eight tetrahedral interstitial sites occupied by fluoride. In fractional coordinates of the conventional cubic unit cell, the calcium basis is $(0,0,0),(0,\frac{1}{2},\frac{1}{2}),(\frac{1}{2},0,\frac{1}{2}),(\frac{1}{2},\frac{1}{2},0)$, and fluoride occupies all combinations $(x,y,z)$ with $x,y,z\in \{\frac{1}{4},\frac{3}{4}\}$. The projection therefore superposes two fluoride sites at each quarter-coordinate position. Labels in the original diagram give fractional depth along $[001]$; periodic boundary sites are displayed on both edges.

Counting shared sites gives $8/8+6/2=4$ calcium sites and eight interior fluoride sites: four formula units per conventional cell.