Find dBm and watts of the attenuator?
There is no need to convert from dB to Watts and back again.  This can be done, and is in fact easier, to do everything in dB.  You can then convert to Watts at the end.

However, the question is ambiguous, at least as you have transcribed it.  It is not clear what "The second attenuator has 20 dB output" is supposed to mean.  If it's a attenuator, then presumably it's passive.  In that case it can't increase the power coming out relative to what is coming in.  It can't have a gain of 20 dB.  Perhaps it is supposed to say a gain of -20 dB, or maybe it is supposed to say that the output is at 20 dBm, but just "20 dB output" in this context makes no sense.

Note that the first "attenuator" does nothing, since it has a gain of 1, which can also be expressed as 0 dB.  Let's assume the second attenuator actually has a gain of -20 dB.  The output signal is therefore:

&nbsp;&nbsp;(40 dBm) + (0 dB) + (-20 dB) = 20 dBm = 100 mW