In your question, you assumed that the "flags" were hexadecimal strings, but I didn’t see anything stating that. They are likely just 30 ASCII characters each, which means they represent 30 raw bytes, not hex. This changes how you calculate the key and IV.

The instructions suggest using the first 8 bytes of each flag to build the key, and the last 8 bytes of each flag to build the IV. Following this logic, I got:

Key: 02c1ef508796f06789510d845b7c1a98

IV: 4d2d1014e8d2dbc92c43d04083d12b9c

Using that key and IV, I was able to decrypt the message like this:
echo "6dU2tgevONWUv6ZWu+84g7E4r4dKOfBxRiY3jnMf2m1aE4r1AZcOztzEKtwve2z211vOnoiXWJTGWTG6wQxibFDw+tVI8hAGwQMqYqeG963g+wz2ppMP+byEcvAgfwvmLrsgm/+nLFxCeKLWYy/e625RmmNEU06s1Dz6izYXX1PNiYn+JAcZQnS1N5KiuvjX1u2qWAIkAPY2H5/BO25vEg==" | base64 -d > cipher.bin

openssl enc -d -aes-256-cbc -in cipher.bin -out decryp.txt -K 02c1ef508796f06789510d845b7c1a98 -iv 4d2d1014e8d2dbc92c43d04083d12b9c

Result: Congratulations! You have completed our first challenge. The final code is 976f01ec317fd664e34ab18a360a43f7888e9065. Please, send it back to us.