Answer
.274 Liters are lost.
Work Step by Step
We find the change in volume of the can and the gas to find:
$\Delta V_{can}=\beta V\Delta T=(3)(12\times10^{-6})(20)(15)=.0108L$
$\Delta V_{gas}=\beta V\Delta T=(95\times10^{-5})(20)(15)=.285L$
Thus, .274 Liters are lost.
