Recall that sound intensity is measured by the formula $d B=10 \log \left(\frac{I}{I_{0}}\right)$. The dripping of a faucet is 126 times as intense as the threshold of sound. How many decibels is the dripping of a faucet? Round your answer to the nearest whole decibel.

Step 1 ：Recall that sound intensity is measured by the formula \(d B=10 \log \left(\frac{I}{I_{0}}\right)\). The dripping of a faucet is 126 times as intense as the threshold of sound.

Step 2 ：The question is asking for the decibel level of the dripping faucet. We know that the intensity of the dripping faucet is 126 times the threshold of sound. We can use the formula for sound intensity to find the decibel level.

Step 3 ：The formula is \(d B=10 \log \left(\frac{I}{I_{0}}\right)\), where \(I\) is the intensity of the sound and \(I_{0}\) is the threshold of sound. In this case, \(I\) is 126 times \(I_{0}\), so we can substitute \(126I_{0}\) for \(I\) in the formula.

Step 4 ：Substituting the values into the formula, we get \(dB = 10 \log \left(\frac{126I_{0}}{I_{0}}\right)\).

Step 5 ：Solving the equation, we get \(dB = 21.00370545117563\).

Step 6 ：However, the question asks for the answer to be rounded to the nearest whole decibel. So, rounding off the above value, we get \(dB = 21\).

Step 7 ：Final Answer: The dripping of a faucet is approximately \(\boxed{21}\) decibels.