If you assume constant-accelerated motion, v^2 = v0^2 + 2ad
200 km/h = 55.6 m/sec
Using the second formula with v = 55.6 m/sec, a = 9.8 m/sec^2 and initial velocity v0 = 0 m/sec gives
d = v^2/(2a) = 158 meters. This is the distance you would free-fall to reach a speed of 55.6 m/sec, assuming constant acceleration of 1 g.
But of course if you are reaching a terminal velocity, then it isn't simply constant acceleration any more. The acceleration is gradually decreasing. You have to solve a differential equation to get the more accurate model of velocity vs. distance or time, and I'm not sure there's a simple formula for the solution. If I were doing this, I'd use a computer.
You could model it on Excel, maybe that will help.
Edited: OK, here's how to do it in Excel.
1. Make four columns for time, distance, velocity and acceleration. I'll assume they're columns A, B, C and D.
2. For initial values put 0, 0, 0, and 9.8
3. In the time column, make even steps in some small increment (0.1 second ought to do) up to say 10 seconds.
4. In the acceleration column, put 9.8 - 0.00317*(previous velocity)^2. That second term accounts for air resistance. That coefficient makes the air resistance = gravity when v = 55.6 m/sec.
5. In the velocity column put v = previous velocity + acceleration * delta-t where delta-t is your time step.
6. In the distance column, put distance = previous distance + velocity * delta-t.
I just wrote a spreadsheet like this. I found that under this model after 10 seconds you've fallen 350 meters, you're moving at 52.5 m/sec and you're still accelerating at about 0.1 g.
At 30 seconds you've gone about 460 meters, you're moving at 54.1 m/sec and your acceleration is down to about 0.05 g.
"Terminal velocity" is an asymptotic value. You don't ever actually get there. So I'm not sure what you'd accept as the answer.