Collision-resistance of $H\Bigl(\bigl(H(m_1\mathbin\| 1)\oplus H(m_2\mathbin\|2)\oplus\dots\oplus H(m_n\mathbin\|n)\bigr)\mathbin\|n\Bigr)$? In this answer and discussion, it is considered a distributable hash

$$\tilde H(m)\ =\ H\Bigl(\bigl(H(m_1\mathbin\| 1)\oplus H(m_2\mathbin\|2)\oplus\dots\oplus H(m_n\mathbin\|n)\bigr)\mathbin\|n\Bigr)$$

where the $m_i$ are $n$ segments (say of fixed size) of message $m$, and $H$ is a cryptographic hash (say of $b$-bit and assumed secure in the ROM).

What can we say about the collision-resistance of $\tilde H$ as a function of $b$ ?
What can we say about the collision-resistance of $\tilde H$ as a function of $b$ ?


That it is effectively nonexistent; it is easy to generate collisions (and second preimages).

You can generate a second preimage by selecting two blocks $m'_i, m''_i$ to correspond to each original block $m_i$; you compute $H(m'_i \mathbin\| i), H(m''_i \mathbin\| i)$, and then use linear algebra (e.g. Gaussian Elimination) to see which combination of $m'_i, m''_i$ all xor the same as the original sequence of $m_i$; that combination hashes to the same value.  As long as the original message is at least $b$ blocks long (that is, $n \ge b$), such a combination exists with good probability.

Once you can create second preimages, you can generate collisions by picking a message and generating a second preimage for it.