No, it cannot be computable. It's a theorem of Tennenbaum from the 50's that there is no computable, non-standard model of Peano arithmetic. If there were a computable model of ZFC, then it would give a computable, non-standard model of Peano arithmetic. Specifically, it would contain some non-standard model for PA, specified by a set of elements and two relations, and the computability of the model of ZFC would imply computability there. (If, say, you want the sum of elements a and b in the model of PA, search by brute force for an element c of the model such that (a,b,c) is in the sum relation. This may be terribly slow, but it will eventually terminate.)
Edit: That's a good point about the computability of (a,b,c), and I'm not sure how to compute that. Fortunately, it turns out not to be needed here. Specifically, if we define (a,b,c) = ((a,b),c) and define (u,v) = {{u}, {u,v}}, then even though it's not clear whether you can computably produce (a,b,c) given a, b, and c, given something you know a priori is a triple, you can try to check whether it comes from a,b,c by finding appropriate elements. (In the simpler case of pairs, to check whether w = (u,v) given that it is some ordered pair, we just need to find x and y such that both are in w, u is in x, and u and v are in y.) Now we just do major brute force: not only searching over all possible choices of c, but also over the auxiliary elements that would establish that (a,b,c) is in the relation. I hope there's a nicer way to do this, but I think it works.
