Here, we can assume that this is a Class C address and not bother about the first three blocks of the IP.
Since we need 20 hosts here, we'll need five bits reserved for the host ID. This is because $2^5$ is the smallest number which can accommodate 20 hosts.
Now, we need to create five subnets - for which we'll borrow 3 bits from the host ID. This is because each borrowed bit divides the network into 2 parts, so 3 borrowed bits will divide it into $2^3 = 8$ parts.
Hence, we are able to create 8 subnets which contain 32 hosts each.
