B

Step 1: Break down the motion of the ball into horizontal and vertical components.
The initial vertical component of velocity (V_y) can be calculated using:
\[ V_y = V \sin(\theta) = 30 \sin(30^\circ) = 30 \times 0.5 = 15 \text{ m/s} \]
Step 2: Use the equation of motion in the vertical direction to find the time (t) taken to reach the ground (height = 20 m):
\[ h = V_y t + \frac{1}{2} g t^2 \]
where g = 9.81 m/s^2.
Plugging in the values:
\[ 20 = 15t + \frac{1}{2} (9.81) t^2 \]
This is a quadratic equation in the standard form:
\[ \frac{1}{2} (9.81)t^2 + 15t - 20 = 0 \]
Step 3: Solving this quadratic equation gives us an approximate time of 2.59 s (not 3 s). Since the options incorrectly state that the ball returns to a height of 20 m again after 3 s, this option is incorrect.
Therefore, the correct answer to the question is B.