For this case we have the following equation:
![l(dB) = 10log( (l)/(lo) )](https://img.qammunity.org/2019/formulas/mathematics/high-school/618xlxg9miq3utbjigdm4c1t2kaw332dzr.png)
We must replace the following value in the equation:
![l = 10^8lo](https://img.qammunity.org/2019/formulas/mathematics/high-school/ytzgz4w104420zealc72n9b46522bi728h.png)
Substituting we have:
![l(dB) = 10log( (10^8lo)/(lo) )](https://img.qammunity.org/2019/formulas/mathematics/high-school/86blh1gnmgtc9ch00prygn6m5e4ftmaly7.png)
Simplifying the given expression we have:
![l(dB) = 10log(10^8)](https://img.qammunity.org/2019/formulas/mathematics/high-school/nm8d5gagjakz4gee3pigjac09ddwg843u0.png)
Then, using logarithm properties in base 10, we can rewrite the expression:
![l(dB) = 10(8)](https://img.qammunity.org/2019/formulas/mathematics/high-school/v2d6sb65d2dkvj1fjpu3ikr0uttxukcsc4.png)
Finally, making the product, the result is:
Answer:
option 4