In n element max heap, smallest element is one of the leaf.. in array representation of max heap, leaves index start from floor(n/2)+1 to n. And there will be max n/2 elements present at last level (height 0).
Just do a for loop from floor(n/2)+1 to n
In which keep track of smallest element
There will be O((n/2)-1) = O(n) comparisons ( we need only 1 element to find)..
Answer =A