Answer: 1022
Class B Subnet Mask given = $255.255.119.0$ = $11111111.11111111.01110111.00000000$
Here it is a non-contiguous subnet mask, which is possible only theoretically, not practically.
In this type of SM, we calculate only no. of $1’s$ as part of NID + SID and $0’s$ as HID.
Thus here, no. of $1’s$ = $22 \ bits$
NID = Class B = $16 \ bits$
SID = $22-16 = 6 \ bits$
HID = $32-22 = 10 \ bits$
Thus no. of hosts/subnet possible = $2^{10}-2 = 1022$