What is the sound level in dB for a sound whose intensity is $5.0 \times 10^{-6}$ $\dfrac{watts}{m^2}$?

Hint: In order to solve the question, we will use the relation between sound level, intensity given and the intensity constant in the processing of solving the answer we will use the logarithmic properties in order to make the calculation easy and hence we will reach to the answer
Formula required to solve the question
$b = 10 \times \log \left( {\dfrac{I}{{{I_o}}}} \right)$
b is represented as sound level
${I_o}$ is represented as intensity constant in watts per meter square
I is represented as given intensity

Complete answer:
In the question we are given the intensity of the sound and we have to find the sound level of that sound in decibel or dB
Intensity of sound (I) = $5.0 \times {10^{ - 6}}{\text{ watt }}{{\text{m}}^{ - 2}}$
In order to find the sound level, we will use the formula
$b = 10 \times \log \left( {\dfrac{I}{{{I_o}}}} \right)$
According to question we have intensity and ${I_o}$ is a constant in the formula of sound level
${I_o}$ = ${10^{ - 12}}{\text{ watt }}{{\text{m}}^{ - 2}}$
Now we will substitute the value of ${I_o}$ and I in the formula of sound level
$b = 10 \times \log \left( {\dfrac{{5.0 \times {{10}^{ - 6}}{\text{ watt }}{{\text{m}}^{ - 2}}}}{{{{10}^{ - 12}}{\text{ watt }}{{\text{m}}^{ - 2}}}}} \right)$
Solving the bracket of log, we get in which ${\text{watt }}{{\text{m}}^{ - 2}}$ cuts in numerator and denominator
$b = 10 \times \log \left( {5.0 \times {{10}^6}} \right)$
Now we will solve the log using the properties of logarithm
$\log \left( {5.0 \times {{10}^6}} \right) = \log (5) + \log ({10^6})$ (Using: ${\text{log}_{\text{a}}}\left( {{\text{uv}}} \right){\text{ = log}_{\text{a}}}{\text{u + log}_{\text{a}}}{\text{v}}$)
 $\log \left( {5.0 \times {{10}^6}} \right) = \log (5) + 6$ (Using: ${\text{log}_{\text{a}}}\left( {{{10}^b}} \right){\text{ = b}}$ )
Now we will substitute the value of $log 5$
$\log \left( {5.0 \times {{10}^6}} \right) = 0.7 + 6$ ($log5= 0.7$)
$\log \left( {5.0 \times {{10}^6}} \right) = 6.7$
Now we will substitute the value of log in the formula of sound level
$b = 10 \times 6.7 = 67{\text{ dB}}$
sound level is equal to $67$ dB

Note:
Many people may confuse how the log ten raised to the power is equal to the number that is raised to the power this is possible as $log 10$ is equal to one so if we use the property of logarithm we will reach to the result as only the number that is raised to the power.