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).