8.
no of cable connected to the host = 90 which contain 2^7 no of bit which is fixed {2^6 < 90 so took 2^7}
as given it is class B address so 255.255.xxxxxxxx.xxxxxxxx
255.255.11111110.0000000 {in 3rd octant 7 bit fixed for subnet) now highlight bit use for host
which contain 2^9 bit = 512 but in this address which is all zero {network address} and last address all 1’s which is broadcast add.
so after removing them 512-2 = 510