sqrt of dB number If the variance is $\sigma^2= 8 {\bf \, dB}$, and I want to calculate standard deviation $\sigma=?$ in dB. Which of the following is correct
a. $\sigma = 8^{0.5} = 2.82 $ dB
b.  $\sigma =10\log_{10}( \sqrt{10^{0.8}})= 4$ dB
 A: I'll go first on a tangent to (properly this time) address the very good point made by lulu in the comments. The decibel is not a proper unit. It's only a signal that the associated numerical value is based on a (specific) logarithmic scale.
Imagine you were a budding scientist 200 years ago. You make measurements of something that feels like "power". Let's call the unit PU. Your measurements have a very high dynamic range. You take notes on pen and paper, and you realize that even your previous neat trick of using powers of $10$ isn't good or convenient enough, because you can't remember if you meant $10^4$ or $104$. That's the point where you start using the decibel logarithmic transformation.
$$
f_{dB}(x) = 10\log_{10}(x) \quad\text{and}\quad f_{dB}^{-1}(x)=10^{x/10}
$$
And so, your $1000$ PU become $30$ dB$_{(PU)}$. Later, you measure standard deviations and variances, and they are respectively in pseudo-units of dB$_{(PU)}$ and dB$_{(PU^2)}$, but rapidly in the literature, everything is just dB, and everyone is supposed to know which underlying hidden unit is being used, and to which dimension.
In conclusion, as Ninad correctly pointed out in the comments, you must always do the $f \circ g \circ f^{-1}$ "unwrapping/rewrapping" when dealing with decibels. So, definitely option b. Yes, there are circumstances where you can take a shortcut. To give a stupid example, you could technically calculate a momentum measured in dB$_{(\text{kg.m.s$^{-1}$})}$ by just adding a mass in dB$_{(\text{kg})}$ and a velocity measured in dB$_{(\text{m.s$^{-1}$})}$. But the OP's case is clearly not one of those.
