Step 1: Recall the FCC unit cell structure.
In a face-centred cubic (fcc) lattice, atoms sit at all 8 corners and at the centre of each of the 6 faces. Understanding where the interstitial voids fall is the key to this question.
Step 2: Locate the octahedral voids.
Octahedral voids in an fcc lattice are found at two sets of positions: (a) the body centre of the unit cell (1 void), and (b) the centre of each of the 12 edges, shared among 4 unit cells each, giving $\frac{12}{4} = 3$ additional voids, for a total of 4 octahedral voids per unit cell. So octahedral voids are present at the body centre AND at edge centres.
Step 3: Locate the tetrahedral voids.
Tetrahedral voids are located at 8 positions inside the unit cell, specifically along the body diagonals (not at edge centres). There are 8 tetrahedral voids per unit cell.
Step 4: Count and compare the voids.
Number of octahedral voids per unit cell $= 4$. Number of tetrahedral voids per unit cell $= 8$. So tetrahedral voids are twice as many as octahedral voids, meaning they are NOT equal.
Step 5: Evaluate the packing efficiency.
Both fcc (cubic close packing) and hcp (hexagonal close packing) have the same packing efficiency of approximately $74\%$. So neither is higher than the other.
Step 6: Identify the correct statement.
Option 1 is wrong (not equal in number). Option 2 is wrong (tetrahedral voids are not at edge centres). Option 4 is wrong (same packing efficiency). Option 3 states octahedral voids are at the body centre and edge centres, which matches our analysis exactly.
\[ \boxed{\text{Option 3: Octahedral voids at body center and edge centers}} \]