1. Notice that the flag is in the _padding_ for the plaintext is the flag we want, not some AES encoded string.
2. Google "chosen plaintext attack aes ecb", come across https://cedricvanrompay.gitlab.io/cryptopals/challenges/09-to-13.html#Challenge-12---Byte-at-a-time-ECB-decryption-(Simple)
3. Implement the attack, retrieve the flag byte by byte: `flag{b4d_bl0cks_for_g0nks}`

Original writeup (https://gist.github.com/AshishMahto/24569356b9f4d83bac6fb10b7fb9f607).