1. **Stating the problem:** We are given the volume change rate of a balloon as $\frac{dV}{dt} = 1.08\pi$ cm³/s and asked to find the rate of change of the radius $\frac{dr}{dt}$ when the radius $r = 3$ cm. 2. **Formula used:** The volume $V$ of a sphere is given by: $$V = \frac{4}{3}\pi r^3$$ 3. **Differentiate both sides with respect to time $t$:** $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ 4. **Substitute the known values:** $$1.08\pi = 4\pi (3)^2 \frac{dr}{dt}$$ 5. **Simplify:** $$1.08\pi = 4\pi \times 9 \times \frac{dr}{dt} = 36\pi \frac{dr}{dt}$$ 6. **Divide both sides by $36\pi$ to solve for $\frac{dr}{dt}$:** $$\frac{dr}{dt} = \frac{1.08\pi}{36\pi} = \frac{1.08}{36} = 0.03$$ 7. **Interpretation:** The radius increases at a rate of $0.03$ cm/s when $r=3$ cm. **Final answer:** $$\boxed{0.03\text{ cm/s}}$$ This corresponds to option c.