In a BST we can't say anything regarding the order of elements that are on separate sides of a node (e.g. 4 and 8, because they are on separate sides of node 6), however, we know that if a node is a parent of another node, it must have been inserted before it.
So 6 is always going to be the first element. Then either 4 or 7 can come (two choices). Then if we look at the remaining structure, 4/2 and 7/8 have exactly the same structure, so we can analyze just one.
Without loss of generality, assume we picked 4. Afterwards we can choose either 2 or 7 (two choices).
If we chose 2, no choice is left. However if we chose 7 either 2 or 8 can come next (two choices). Then we're left with no choice.
So the total number of permutations is $2(1 + 2) = 6$. The permutations are: