A subnetted Class B network has the following broadcast address :…

Key idea: determine which subnet block sizes (in the third octet) can produce a broadcast ending at 95.255 for a Class B network.

• 255.255.224.0 (/19): block size = 32 in the third octet → blocks are 0–31, 32–63, 64–95, ... ; the block 64–95 has broadcast 144.16.95.255.

• 255.255.240.0 (/20): block size = 16 in the third octet → blocks include 80–95 ; that block’s broadcast is 144.16.95.255.

• 255.255.248.0 (/21): block size = 8 in the third octet → blocks include 88–95 ; that block’s broadcast is 144.16.95.255.

Conclusion: Since the broadcast 144.16.95.255 can be produced by subnets under any of the three masks, the correct answer is that the mask could be any one of 255.255.224.0, 255.255.240.0, or 255.255.248.0. Options that claim a single mask is "necessarily" the one are incorrect.

Quick method to check similar problems: identify the class to know the base octet boundary, determine the block size from the subnet mask (256 − value of masked octet increment), and see whether the given address falls on a subnet boundary or broadcast boundary for that block size.