Reusing Todd Lammle's method (CCNA) here, which I used in another answer.
$/27$ bits turned on is $255.255.255.224$. Thus $256 - 224 = 32$ is the block size for the last octet. So subnet addresses follow the pattern $18.104.22.168$, $22.214.171.124$, $126.96.36.199$, $188.8.131.52$, $184.108.40.206$, $220.127.116.11$ and so on.
Thus we can easily see that the given IP address is in the $18.104.22.168$ subnet, whose last valid host address is $22.214.171.124$. (as $126.96.36.199$ is the broadcast address).