Can I have a hint on this combinatorics question?

Two cubes that differ by a rotation are considered the same. There are $24$ ways to orient the cube, all of which are distinct because of the use of distinct colours, so $6!$ must be divided by $24$ to get the correct answer of $30$.

This is perhaps a more concrete way to arrive at the answer. Say we will color the faces with the colors, $R,O,Y,G,B,V$. We have to color some face $R$. Place the cube on the table, with that face down. Now we have to color some face $O$.

We can paint the top face, or one of the lateral faces. Suppose first that we paint the top face. Now we have to paint one of the lateral faces $Y$. Whichever face we choose, we can rotate the cube so that that face is facing us. Now the cube is fixed in place. $R$ on bottom, $O$ on top, $Y$ facing us. We can't rotate it without changing the arrangement. There are $3!$ ways to paint the remaining $3$ faces.

If instead, we paint one of the lateral faces $O$, we can rotate the cube so $O$ is facing us, and again we find it is fixed in place; $R$ on bottom, $O$ facing us. We have $4!$ ways to paint the remaining $4$ faces.

You may be interpreting the question a bit too literally. There are indeed $6!$ orders in which we can choose to color the faces of the cube, but in many cases, the results are indistinguishable. The question is asking for the number of distinct cubes that can be made by coloring each face with one of $6$ given colors.