The average deceleration of the speed boat is given by:

$$a=\frac{\Delta v}{\Delta t}=\frac{0\text{ m/s}-25\text{ m/s}}{3\times60\text{ s}}=\frac{-25}{180}\text{ m/s}^2=\boxed{-0.139\text{ m/s}^2}$$

Therefore, the rate of deceleration of the speed boat is 0.139 m/s^2.