SNR and Attenuation Question
If you are new to these types of problems, I would recommend converting all original dB terms to straight power ratios and then converting back to dB form in the end.

Start with the original SNR of 25 dB. To convert this to a power ratio, use 10 (25/10) which is 316.2. This means the signal contains 316.2 times more power than the noise.

Convert the 10 dB power loss figure to a power ratio the same way to find it is 10.

Now you can walk through the gains and losses as straight multiplication.  The signal goes down by a factor of 10 so it is now 31.62. The noise doubles so it goes up to 2 (it was the 1 denominator of the 316.2).

So we have a new SNR of 31.62/2 or 15.81. Convert this back to dB by the formula 10*log 10 (power ratio) and you find that the new SNR is ~ 12 dB.