Class B network has the form N: N: H: H, the default subnet mask is 16 bits long.
There are an additional 7 bits to the default subnet mask. the total number of bits in
the subnet are 16+7 = 23.
This leaves us with 32-23 = 9 bits for assigning to hosts.
7 bits of subnet mask corresponds to 2^7-2 = 128-2 = 126 subnets.
9 bits belonging to host addresses correspond to 2^9-2 = 512-2 = 510 hosts.
so, the answer will be 510 hosts and 128 subnets