Can minimum or maximum height of the binary search tree be constrained by the position of some elements
The question you were asked talks about 4 levels, which is a height of 3, not a height of 4.

Answer 1 is incorrect because there is a tree of height 3 in which G is at the root (you provided an example already).

Answer 3 is incorrect because of the following tree:

         F
       /  \
      /    \
     /      \
    B        J
   / \      / \
  A   D    H   L
     / \  / \ / \ 
    C   E G I K M  


Answer 4 is incorrect because of the following tree:

           G
         /  \
        /    \
       /      \
      D        K
     / \      / \
    B   F    I   M
   /\  /    / \ / 
  A  C E    H J L


Answer 2 is, as you answered correctly, the only true one. 

To see this: Assume T is a tree which satisfies the constraints, i.e.

           1
         /  \
       /      \
     /          \
    B            K
  /   \        /   \
  2   3       4     5
 / \ / \     / \   /  \
6  7 8 9   10  11  12 13


Where the numbers 1 to 13 represent unknowns from A,C,D,E,F,G,H,I,J,L,M and two empty symbols.

As you can see, both the left and the right subtree can contain at most 6 elements. Thus, the root has to be $1 \in \{F,G,H\}$.

If $1=F$, then the right sub-tree has to contain {G, H, I, J, K, L, M}. However, K is the root of the right sub-tree and therefore $\{4,10,11\} = \{G,H,I,J\}$, which is impossible. Using symmetrical arguments we can show that the root cannot be H either.

Therefore it must be $1 = G$. But then $\{3,8,9\} = \{C,D,E,F\}$ which is a contradiction as well.