Alter plaintext through IV in block cipher AES in CBC mode
Each bit you flipped in the IV changes the same bit in the decrypted plaintext. So if you want to change the first E character of ENCRYPTED into a T of THE then you need to XOR the binary representation of E and T (probably ASCII) and XOR that with the byte in the IV. This will of course only work for the first block of 16 bytes. Fortunately "SEND THE DATA" is small enough for that (13 bytes if I count correctly).

That leaves the problem of the padding. The original block contains the padding in the second block because of the size (19 bytes). However, you just have to adjust the padding of the first block by performing the XOR as displayed above. Instead of XOR'ing the final 16 - 3 = 3 plaintext bytes with a letter you XOR them with the value 03 in hex and XOR that with the IV. Now you've also generated the required padding.

Finally, you need to leave out the second block. And that's all me wrote, happy programming.