Let us assume that the balloon C continues to rise at an acceleration of "a" = 2 m/s² after the 1st stone is dropped at t =0 sec. The first stone A has the same initial velocity upwards as the 2nd stone B at t = 0 sec. So relative velocity between the two stones is 0.
Acceleration of B after t = 0, is = a = 2 m/s² wrt ground
Acceleration of A after t = 0, is = -g = -10m/s² wrt ground.
Relative to 2nd stone B acceleration of A = (a+g)
In 2 seconds, the separation between A and B = s
s = ut+1/2 at² = 0 * 2 + 1/2 (g+a) 2² = 1/2 (2+10)2² = 24 meters
Alternately, we can assume an initial velocity u of the balloon at t =0, and the stone is dropped at t = t1 and the second stone is dropped at t1 + 2 sec.
Vertical position of stone 1 at t1+2 sec = ( u t1+1/2 a t1² ) + (u + a t1) * 2 - 1/2 g 2²
Vertical position of stone 2 at t1+2 sec = u (t1 + 2) + 1/2 a (t1+2)²
Difference between them = 1/2 (a+g) 2² = 1/2 * 12 * 4 = 24 meters