What is necessary and/or sufficient requirement for a subring of a field to be computable? As title asks, what is necessary and/or sufficient requirement for a subring of a field to be a computable ring?

Conditions on either field or subring are fine.
The question of how to find computable substructures of algebraic structures was studied by Jens Blanck and myself in the paper "Canonical Effective Subalgebras of Classical Algebras as Constructive Metric Completions". There we give general conditions on what it means for a substructure of an algebraic structure to be computable. Let me give a summary, but I recommend reading the paper for thorough understanding.

We work with respect to an arbitrary model of computation, such as Turing machines, Type II Turing machines, topological models, $\lambda$-calculus, a programming language, etc. The formalism which encompasses all of these is realizability theory.

There is an embedding $\nabla : \mathsf{Set} \to \mathsf{Asm}$ from the category of sets to the category of assemblies (over some partial combinatory algebra, see our paper). The category of assemblies can be thought of as a category of sets equipped with a computability structure. The embedding $\nabla$ maps sets to sets equipped with trivial computability structure, namely, if $S$ is a set then $\nabla S$ is the set $S$ in which every $x \in S$ is realized by every realizer (if you've never heard of realizability before, think of Gödel encodings of objects). Intuitively this says that when we look at a realizer we have absolutely no idea which element it represents.

Assume a fixed signature $\Sigma$ for an algebraic structure (such as one of a ring) and let $\mathcal{A}$ be a $\Sigma$-algebra in the category of sets, i.e., what people normally think of when they say $\Sigma$-algebra. Because $\nabla$ preserves products, it maps $\mathcal{A}$ to a $\Sigma$-algebra $\nabla \mathcal{A}$ in $\mathsf{Asm}$. Now, a computable $\Sigma$-subalgebra of $\mathcal{A}$ is just a subalgebra of $\nabla \mathcal{A}$ in the category $\mathsf{Asm}$.

When we unravel what this means in terms of realizers we see that it is the expected thing: for $\mathcal{B}$ to be a computable subalgebra of $\mathcal{A}$ it has to be a subalgebra in the usual sense (of course), you have to give a realizability relation for the carrier of $\mathcal{B}$ which explains how the elements of carrier are realized, and all the operations on $\mathcal{B}$ must be realized (computable) with respect to the chosen realizability relation on the carrier.

It occurs to me that maybe you are asking the following question: given a subalgebra $\mathcal{B}$ of an algebra $\mathcal{A}$ is there some computable structure on $\mathcal{B}$, i.e., is there some way of making all the operations on $\mathcal{B}$ computable? First note that $\mathcal{A}$ does not play a role here. Second, if the only requirement is that the operations on $\mathcal{B}$ be computable, then the answer is always "yes, just use the trivial computability structure $\nabla \mathcal{B}$". That is, represent every element of $\mathcal{B}$ with any realizer you want, say 42, and then all operations are implemented by functions that always return 42. In order to avoid such a silly answer you have to impose further conditions -- and that's what our paper is about.