1. **State the problem:** The intensity of a sound wave, $I$, is doubled. We need to find the difference in sound level, $\Delta L$, measured in Decibels (dB). 2. **Formula used:** The sound level $L$ in decibels is given by: $$L = 10 \log_{10}\left(\frac{I}{I_0}\right)$$ where $I_0$ is the reference intensity. 3. **Difference in sound level:** When intensity changes from $I$ to $2I$, the difference in sound level is: $$\Delta L = 10 \log_{10}\left(\frac{2I}{I_0}\right) - 10 \log_{10}\left(\frac{I}{I_0}\right)$$ 4. **Simplify the expression:** $$\Delta L = 10 \left[\log_{10}(2I) - \log_{10}(I)\right] = 10 \log_{10}\left(\frac{2I}{I}\right) = 10 \log_{10}(2)$$ 5. **Evaluate $\log_{10}(2)$:** $$\log_{10}(2) \approx 0.3010$$ 6. **Calculate $\Delta L$:** $$\Delta L = 10 \times 0.3010 = 3.010$$ **Final answer:** The difference in sound level is approximately $3.01$ decibels.