If I understood correctly, instead of storing the full hash value, you just store 2-byte. As always, we assume that the attacker has the random chunk information, too.

It will be very easy to find a collision since your truncated hash contains only $2$-byte $= 16$-bit. By the birthday attack on hash functions, the attacker only needs to try $\sqrt{2^{16}}= 2^8 = 256$ different files to has a collision with 50%.

you can also see this question and answers for truncated hash values;

• What are the consequences of removing a single byte from a sha256 hash?
• Is truncating a SHA512 hash to the first 160 bits as secure as using SHA1?

If I understood correctly, instead of storing the full hash value, you just store 2-byte. As always, we assume that the attacker has the random chunk information, too.

It will be very easy to find a collision since your truncated hash contains only $2$-byte $= 16$-bit. By the birthday attack on hash functions, the attacker only needs to try $\sqrt{2^{16}}= 2^8 = 256$ different files to has a collision with 50% probability.

You can also see this questions and answers for truncated hash values;

• What are the consequences of removing a single byte from a sha256 hash?
• Is truncating a SHA512 hash to the first 160 bits as secure as using SHA1?

If I understood correctly, instead of storing the full hash value, you just store 2-byte. As always, we assume that the attacker has the random chunk information, too.

It will be very easy to find a collision since your truncated hash only contains 2-bytes $= 16$-bit. By the birthday attack on hash functions, the attacker only needs to try $\sqrt{2^{16}}= 2^8 = 256$ different files to have a collision with 50%.

you can also see:

• What are the consequences of removing a single byte from a sha256 hash?
• Is truncating a SHA512 hash to the first 160 bits as secure as using SHA1?

If I understood correctly, instead of storing the full hash value, you just store 2-byte. As always, we assume that the attacker has the random chunk information, too.

It will be very easy to find a collision since your truncated hash contains only $2$-byte $= 16$-bit. By the birthday attack on hash functions, the attacker only needs to try $\sqrt{2^{16}}= 2^8 = 256$ different files to has a collision with 50%.

you can also see this question and answers for truncated hash values;

• What are the consequences of removing a single byte from a sha256 hash?
• Is truncating a SHA512 hash to the first 160 bits as secure as using SHA1?