This mini RSA question is not like previously seen ones where we could simply apply a cube root attack. Due to padding, m^e is actually larger than the public modulus.
Although we cannot run a cube root attack, we do realize that e is only 3 and m^e is barely larger than n we could see brute force the plaintext by just trying to add the public modulus.
RSA encryption uses : c = m^e mod n
RSA decryption could use: p = (c+xn)^(1/e)
Here, x is probably not so large so we can test that theory out. Initially, I used the decimal python module but the results were not accurate enough thus not giving me a flag. I move on to use a sage math kernel which handles numbers way better than python and it gave me the flag with ease.