BFS can also be used to find the depth of a binary tree.
The depth of node is defined as the number of edges from the root to the node.
They have given that $n^{th}$ vertex is found at depth 4,what is the maximum value of n?
At each depth, we can have maximum of $2^d$ node where d=current depth.
At, d=4, we would have $2^4=16$ nodes and assuming our tree is complete binary tree, The number of node till depth 3 are
$2^0+2^1+2^2+2^3=15$ nodes.
So, if root starts from as node 1, till depth 3, we would have node numbered 15.
At depth 4, we have 16 nodes and they will be numbered till 16+(16-1)=31. i.e. node numbered from 16 to 31.
So, maximum numbered node our BFS can see at depth 4 is 31.
Answer-31