Consider the following example – Size of Plaintext – 48 bytes. Size of MAC – 20 bytes. Block Length – 8 bytes. How many bytes of padding need to be added to the system?

Correct Answer: 3
48 + 20 = 68 bytes. 72 is the next multiple of 8 (Block Length). 72 – 68 = 4. But we need to compensate 1 byte for length of the padding. Therefore, we require only 3 Bytes padding.