B

Step 1: Understand the situation: The person in the bus perceives the rain to be falling vertically at a velocity of 20 m/s while the bus moves eastward at 10 m/s. This means that the actual velocity of the rain is a combination of its vertical and horizontal components.

Step 2: Let's denote the velocity of the rain with respect to the ground as \( V_r \). We can break this down into two components:
- The horizontal component with respect to the ground: \( V_{rh} \)
- The vertical component with respect to the ground: \( V_{rv} \)

Step 3: Since the rain appears to fall vertically to the person in the bus, we can use the concept of relative velocity. The apparent velocity of the rain is the vector sum of the rain's velocity and the bus's velocity.
So, if we take the vertical motion downwards as positive, we can write the equation in terms of components:
\( V_{rv} = 20 \) m/s (since the rain appears to be falling vertically downward)
and \( V_{rh} = 10 \) m/s (the velocity of the bus, acting opposite to the rain's horizontal component).

Step 4: Now, the actual speed of the rain with respect to the ground is given by the Pythagorean theorem as it forms a right triangle with horizontal and vertical components:
\( V_r = \sqrt{V_{rh}^2 + V_{rv}^2} \)
Plugging in the values:
\( V_r = \sqrt{(10)^2 + (20)^2} = \sqrt{100 + 400} = \sqrt{500} = 10\sqrt{5} \) m/s.

Step 5: Evaluating \( 10\sqrt{5} \):
Since \( \sqrt{5} \approx 2.236\), we find \( 10\sqrt{5} \approx 22.36 \) m/s.

Therefore, the speed of the raindrops with respect to the ground is approximately 22.36 m/s.