Solution:

There are two "bands" of individual unit cubes (called cubies) that are moved by $M$. Of those, only the cubies exactly three units from an edge of the cube, such as the cubie originally at the intersection of the shaded bands, can ever move out of a single plane of rotation. All the other cubes either remain fixed or return to their original position every four repetitions of $M$ as they travel once around the cube with four $90$ degree turns.

Each of these cubies that move in both the planes of rotation will return to its original position every seven repetitions of $M$. It must be moved by a total of four $90$ degree turns around each of $X$ and $Y$ to return to its starting point; to do so, it moves once with both $X$ and $Y$, three times with $X$, and three times with $Y$, in some order depending on the cubie's initial position. For example, the cubie that is initially at the top intersection of the two shaded bands will move first to the right face (where the $Y$ arrow emerges) and then to the bottom and then the left. On the fourth move $M$ it will be affected by first $X$ and then $Y$, ending up on the back. Then it is affected only by $Y$ for three moves, shifting to the bottom, front, and finally returning to its original location on top after seven moves.

Thus, for all the cubies to return to their starting points, the number of moves $M$ must be a multiple of both $4$ and $7$, and therefore $28$ is the smallest possible number of repeats of $M$ in order for the cube to be restored.