Which field is it?

No, $m$ is very far from unique. Take any function $p : \mathbb{N} \to \text{primes}$ and any non-principal ultrafilter $U$ on $\mathbb{N}$. Then $R$ admits a quotient map to $\prod_{i=1}^{\infty} \mathbb{F}_{p(i)}$ which in turn admits a quotient map to the corresponding ultraproduct of the fields $\mathbb{F}_{p(i)}$, which is a field, and this map (viewed as a map with domain $R$) contains $I$ in its kernel. (I think this construction describes all possible $m$ but am not sure.)

Generically these fields are pseudo-finite fields and in particular quasi-finite fields. When they are not themselves isomorphic to finite fields (which happens iff there exists some prime $p$ such that $\{ i : p(i) = p \}$ is contained in the ultrafilter $U$, and in particular if $\{ i : p(i) = p \}$ is cofinite), they are unfamiliar.