the official solution, below quoted, is not correct, so it remains unsolved Quote: Solution. No, it cannot. Let us call a series of faces $F_1, F_2, .. , F_k$ a ring if the pairs $(F_1, F_2), (F_2, F_3),.. , (F_{k-1}, F_k), (F_k, F_1)$ each have a common edge and all these common edges are parallel. It is not difficult to see that any two rings have exactly two common faces and, conversely, each face belongs to exactly two rings. Therefore, if there are n rings then the total number of faces must be $2 \binom{n}{2} = n(n - 1)$. But there is no positive integer n such that $n(n - 1) = 1992$. Remark. The above solution, which is the only one proposed that is known to us, is not correct. For a counterexample, consider a cube with side $2$ built up of four unit cubes, and take the polyhedron with $24$ faces built up of the faces of the unit cubes that face the outside. This polyhedron has rings that do not have any faces in common. Moreover, by subdividing faces into rectangles sufficiently many times, we can obtain a polyhedron with $1992$ faces.

parmenides51 wrote: so it remains unsolved Are you sure? Isn't the last sentence of the remark saying that the answer actually is YES with a very simple construction: Similar to the counterexample, we divide the cube into $2 \times 18 \times 48$ rectangular boxes and obtain a polyhedron with $2(2 \cdot 18+2 \cdot 48+18 \cdot 48)=1992$ rectangular faces (in particular parallelograms).

I do not know anything about Solid Geometry (shame on me, I know). I just quoted everything written in the official solutions. So I cannot understand any of these. I just though it was worth mentioning the comment found there. here is the solutions' pdf