Please give me the solution.

height of point of contact from the table top= h
Circle(O1. 8) & Circle(O2, 2)
Line from O2 touching the vertical radius be at point A
Angle subtend by O1A at O2 be x
And point of contact of the two spheres be B
So, In triangle(O1AO2)
Sin x = perpendicular/hypt.
=6/10=3/5 (i)

In triangle O2BC

Sin x =BC/ O₁B = 3/5 ...(fro eqn i)
So, BC=(O₁B) (3/5)
(2)(3/5)
BC= (6/5)
So,
h = BC + radius of smaller circle
=1.2 + 2
=3.2