To solve this problem, we use the formula for the decibel level:

$D = 10 \cdot \log_{10} \left( \frac{I}{10^{-12}} \right)$

Here, the intensity $I$ is given as $10^{-2} \, \text{W/m}^2$. Substituting $I = 10^{-2}$ into the formula:

$D = 10 \cdot \log_{10} \left( \frac{10^{-2}}{10^{-12}} \right)$

Simplify the fraction:

$\frac{10^{-2}}{10^{-12}} = 10^{-2 + 12} = 10^{10}$

Now calculate the logarithm:

$\log_{10}(10^{10}) = 10$

So the decibel level is:

$D = 10 \cdot 10 = 100 \, \text{dB}$

Final Answer:

The decibel level is 100 dB.

Would you like more explanation or help with similar problems? Here are some follow-up questions:

1. What happens to the decibel level if the intensity increases by a factor of 10?
2. Can you derive the formula for $D$ from first principles?
3. How is sound intensity related to perceived loudness in real-world scenarios?
4. What would the decibel level be for $I = 10^{-4} \, \text{W/m}^2$?
5. Why is $10^{-12} \, \text{W/m}^2$ used as a reference intensity?

Tip: Always double-check the base of the logarithm in formulas, as decibel calculations typically use base 10!