This question concerns block cipher padding.Suppose the block cipher has a block size of 15 bytes.A certain message ends with a partial block which is 4bytes. We must reversibly pad out the message to usesome block cipher modes, such as CBC (even if the lastblock is full).Consider the following padding strategy.Determine the number of padding bytes required. This isa number n which satisfies 1 ≤ n ≤ 15 and n + 1(P)is a multiple of 15, where 1(P) is the length of theunpadded plaintext. Pad the plaintext by appending nbytes, each with value n.Suppose the final (possibly partial) block of the messageis "0x00010203" in hexadecimal. Write out the completefinal block in hexadecimal. Please do not put a leadingOx as this has been written already.Ox

In block cipher padding, we're dealing with a block size of 15 bytes. If a message ends with a partial block of 4 bytes, we need to pad it out for certain block cipher modes like CBC.Here's the detailed padding strategy:Determine the number of padding bytes required (let's call it n). It should be between 1 and 15, and n+1 added to the length of the unpadded plaintext (l(P)) should result in a multiple of 15.Pad the plaintext by appending n bytes, each having the value n.Now, suppose the final (possibly partial) block of the message is "Ox00010203" in hexadecimal. Let's apply the padding:l(P)=4 (length of the partial block) n+1 should be a multiple of 15. The smallest n satisfying this condition is 10 (10+1=11, which is a multiple of 15).Pad the block with 10 bytes, each having the value 10 (since n=10).Hence The complete final block after padding is 000102030a0a0a0a0a0a0a0a0a0aThe complete final block without the leading "0x" is000102030a0a0a0a0a0a0a0a0a0a

Answered: This question concerns block cipher…