Let no of elements in each equivalence class be 'n'.
Now, asper handshaking theorem, there will be nP_{2} elemnts in the relation R corresponding to single equivalence class. Similarly , nP_{2 }for the remaining 2 classes.
So, total elements = (nP_{2})^{3}
Here, in the options, only option (C) is proper cube. So, I think (C) is the answer.