A host address is 192.168.14.222/28. Which address is the broadcast address of its subnet?
With /28, the host portion occupies only the last 4 bits, so the usable range for the subnet containing .222 is 192.168.14.208 through 192.168.14.223. Setting all host bits to 1 yields the broadcast address 192.168.14.223, which is the directed broadcast for the /28 subnet that includes .222. This makes .223 the correct answer.
Why this answer
A /28 subnet has a block size of 16. In practical terms, the fourth-octet blocks are 0-15, 16-31, and so on. Because 222 falls within the 208-223 block, the broadcast address is the last address in that block: 192.168.14.223.
This is a subnet-boundary question that depends on identifying the correct /28 block before choosing the broadcast address.
Exam trap
Be careful not to confuse the broadcast address with the network address of the next subnet or a host address within the subnet.
Why the other options are wrong
192.168.14.207 is the broadcast address of the previous /28 subnet (192.168.14.192/28), not the subnet containing 192.168.14.222.
192.168.14.208 is the network address (subnet ID) of the /28 subnet containing .222, not the broadcast address.
192.168.14.224 is the network address of the next /28 subnet (192.168.14.224/28), not the broadcast of the current subnet.