the 188.8.131.52 IP address, we borrow 2 bits to create subnetworks.
How many hosts can we have in each subnetwork?
Why? Why 256 and not something else?
Based off of normal nth power math coupled to subnetting where you subtract the network and broadcast addresses, a possible answer would be 254 and not 256. To support 256 addresses you would need a /23 which would provide 510 addresses (512 - 2). Since 184.108.40.206 falls within the class C range we automatically know there are 24 network bits. If you borrow 2 bits from the remaining hosts bits you end up with a /26 So the answer that @stryder144 provided is correct. A /26 would support 62 hosts in each subnetwork.
Since there is no slash notation used in the IP address, I will assume it is a class C network. If you borrow 2 of the 8 host bits that are available, you get 6 bits for the hosts. 2 to the 6th power is 64, subtract 2 (one address for the network and one for the broadcast), that leaves you with 62 possible hosts...unless my math and/or logic is wrong.