1st minimum will always at root of min heap. Hence 0 comparisons.
Now 2nd minimum can be child left child or right child of the root. Hence 1 comparison.
For the 3rd minimum we are assuming that 2nd min is the left child of root. So the 3rd minimum must be in the childs of 2nd min or right child of 1st min(root). Because if 3rd min is found other places than these nodes then min heap property will be violated (try yourself by some examples). So total comparison 2.
Now same way I am assuming the 3rd min is the left child of the 2nd min so the 4th min can be found in the the area in the picture below. So to find 4th min we have to compare 4 nodes means 3 comparisons.
You can try the remaining yourself.
1st min → 0 comp
2nd min → 1 comp
3rd min → 2 comp
4th min → 3 comp
5th min → 4 comp
………
……….
kth min → k-1 comp
To finnd kth min we need k*(k-1)/2 comp.
So to find 7th min we need 7*(7-1)/2 = 21 comp.
So time needed $\Theta$(1).
Option: D