Class B has 2 Bytes for the Network ID. The remaining 6 bits will be used for the subnet ID.
So,total number of subnets possible = $2^{6}$
= 64 subnets
Remaining (32-16-6=10) bits will be used for the hosts
So, total number of hosts possible = $2^{10}$
= 1024 hosts
2 of the these host IDs are not available for general purpose (1024-2=1022)
As, none of the option matches
Option (B) is close to the answer.