From the given source, we know that 2 encryptions with 3 signs out of string.printable are performed, which has 100 members. So brute forcing all double-encryption-combinations would take 100^3^2 units of time. Or 10^12.
Luckily wew can try to encrypt the known plaintext `aaaaaaaaaaaaaaaa` and decrypt the known ciphertext `ef92fab38516aa95fdc53c2eb7e8fe1d5e12288fdc9d026e30469f38ca87c305ef92fab38516aa95fdc53c2eb7e8fe1d5e12288fdc9d026e30469f38ca87c305` and if these match, we found a pair of 1st and 2nd keys that result in the correct encryption.
Then we only have to decrypt the flag with the 2nd and then with the 1st.
The keys were `"0000000000000t\r["` and `}=v0000000000000`
# Solver source code```python#!/usr/bin/python3
from string import printableimport itertoolsfrom Crypto.Cipher import AESfrom binascii import hexlify, unhexlify
pt = hexlify(b"aaaaaaaaaaaaaaaa")target = unhexlify(b"ef92fab38516aa95fdc53c2eb7e8fe1d5e12288fdc9d026e30469f38ca87c305ef92fab38516aa95fdc53c2eb7e8fe1d5e12288fdc9d026e30469f38ca87c305")lookup_table = {}
# create all 100^3 forward-encryption-optionsfor p1 in itertools.product(printable, repeat=3):	key = "".join(p1)	key1 = "0000000000000"+key
	cipher1 = AES.new(key=key1.encode("utf-8"), mode=AES.MODE_ECB)	c1 = hexlify(cipher1.encrypt(pt))	lookup_table[c1] = key1
print("done creating lookup table")
# create all 100^3 backward-encryption-optionsfor p2 in itertools.product(printable, repeat=3):	key = "".join(p2)	key2 = key+"0000000000000"
	cipher2 = AES.new(key=key2.encode("utf-8"), mode=AES.MODE_ECB)	c2 = cipher2.decrypt(target)	if c2 in lookup_table: # check if the decryption of target == encryption of known plaintext		key1 = lookup_table[c2]		print(key1, key2)		break
# part 2: decryptionflag = b"fa364f11360cef2550bd9426948af22919f8bdf4903ee561ba3d9b9c7daba4e759268b5b5b4ea2589af3cf4abe6f9ae7e33c84e73a9c1630a25752ad2a984abfbbfaca24f7c0b4313e87e396f2bf5ae56ee99bb03c2ffdf67072e1dc98f9ef691db700d73f85f57ebd84f5c1711a28d1a50787d6e1b5e726bc50db5a3694f576"flag = unhexlify(flag)
for key in [key2, key1]:	cipher = AES.new(key=key.encode("utf-8"), mode=AES.MODE_ECB)	flag = unhexlify(cipher.decrypt(flag))	print(flag)
# key1 = "0000000000000t\r["# key2 = "}=v0000000000000"```