Answer
$8.3\times10^{-9}N$.
Work Step by Step
Momentum is conserved. The light leaving the flashlight gains momentum, given by equation 22–9a. Dividing the change in momentum by the elapsed time tells us the force that the flashlight applies to the emitted light.
$$\frac{\Delta p}{\Delta t}=\frac{\Delta U}{c\Delta t}=\frac{P}{c}$$
$$=\frac{2.5W}{3.00\times10^8 m/s}=8.3\times10^{-9}N $$
By Newton’s third law, this equals the reaction force exerted upon the flashlight.