Correct option (2) 8(√6 + √2 + 2)

O → centre of sphere
P,Q → point of contact of tangents from A
Let T be top most point of balloon & R be foot of perpendicular from O to ground.
From triangle OAP, OA = 16cosec30° = 32
From triangle ABO, OR = OA sin75° = 32 \(\frac{(\sqrt{3} + 1)}{2\sqrt{2}}\)
So level of top most point = OR + OT
= 8\((\sqrt{6} + \sqrt{2} +2)\)