A boy standing on a bridge throws a ball up at the speed of 20m/s, the ball aftr 3 seconds touches water surface below. Can you calculate the height of bridge to be 5m?

Unless the boy is on another planet, the problem as stated makes no sense. With Earth's gravity (a=-9.8m/s^2) being the only influence on the ball, at the end of 3 seconds it will be
  h = v₀t - (1/2)at² = 20(3) - (1/2)(9.8)(3^2) = 15.9
meters above where it was thrown from.

The surface of the water in the aqueduct is 15.9 m above the point of release of the ball. There is not enough information to tell the height of the bridge.