in a class B network 16 bits are for host id part and 16 bit for network id part
so network id of given class B network is 160.24.0.0
broadcast address of one of the subnet of this network is 160.24.223.255
it is directed broadcast address in fact because it has all ones in host id part of the subnet
expanding only hid part to analyze
160.24.11011111.11111111
so if it it DBA then it must have all 1's in HID part of subnet and since it has 13 Least significant bits as 1's we may confirm that hid part has max 13 bits means it may be less than that also.
case 1) : assuming no of bits in hid part = 13 then
16 bits are NID part and 3 bits are for subnet id part hence,
starting 19 bits will be subnet mask (all ones in NID bits and subnet id bits)
so
subnet mask maybe
11111111.11111111.11100000.00000000 (255.255.224.0)
case 2) : assuming no of bits in hid part = 12 then
16 bits are NID part and 4 bits are for subnet id part hence,
starting 20 bits will be subnet mask (all ones in NID bits and subnet id bits)
so
subnet mask maybe
11111111.11111111.11110000.00000000 (255.255.240.0)
similarly we may assume no of bits in hid part anything less than 13
and hence in subnet mask number of 1's will be 19 or more than that
so subnet mask will be anyone in
(255.255.224.0) or (255.255.240.0) or (255.255.248.0) and so on.
so here M2 and M3 are correct.