Step 1: Use the continuity equation.
The continuity equation for an incompressible flow is given by:
\[
A_1 v_1 = A_2 v_2
\]
Where:
- \( A_1 = \frac{\pi d_1^2}{4} \) is the cross-sectional area at the entrance,
- \( A_2 = \frac{\pi d_2^2}{4} \) is the cross-sectional area at the exit,
- \( v_1 = 3 \, \text{m/s} \) is the velocity at the entrance,
- \( d_1 = 0.25 \, \text{m} \) and \( d_2 = 0.20 \, \text{m} \) are the diameters at the entrance and exit, respectively.
Step 2: Solve for \( v_2 \).
Rearranging the continuity equation to solve for \( v_2 \):
\[
v_2 = v_1 \times \frac{A_1}{A_2} = 3 \times \frac{\left( \frac{\pi (0.25)^2}{4} \right)}{\left( \frac{\pi (0.20)^2}{4} \right)} = 3 \times \left(\frac{0.25^2}{0.20^2}\right) = 3 \times \left(\frac{0.0625}{0.04}\right) = 4.68 \, \text{m/s}
\]
Final Answer: \[ \boxed{4.68 \, \text{m/s}} \]

