Step 1: Recall power distribution in AM.
Sideband power $= \dfrac{\mu^2}{2} P_c$, where $\mu$ = modulation index.
Step 2: Apply $\mu = 0.3$.
Percentage power in sidebands = $\dfrac{\mu^2}{2 + \mu^2} \times 100 = \dfrac{0.09}{2.09} \times 100 \approx 43%$.
Final Answer:
\[
\boxed{43%}
\]