Meta-incomputability Is there a set $B$ about which it provably cannot be decided whether it is computable in $\mathsf{ZFC}$?
 A: An explicit example: $B=$ the set of all theorems of ZFC.
A: It's worth noting that the answer above is answering the following question: is there a formula $\phi(x)$ in the language of ZFC such that ZFC can't prove either $\lbrace x \in \omega \mid \phi(x) \rbrace$ is computable nor that it's not computable.  One might worry the answer above is 'cheating' in a certain sense by picking a formula $\phi$ which behaves very differently depending on whether or not CON(ZFC) is true.  
If we wanted to think about sets a bit more extensionally we could insist that our formula not change it's mind about what actual integers are in $B$ depending on the model.  In other words we could interpret the question as asking if there is some definition of a set $B$ whose standard part is the same in all models of $ZFC$ but some models extend to be computable while others extend it to be uncomputable.
The answer here is yes as well.  Let $B$ be the set of $n$ such that $n \in 0'$ (or any other provably uncomputable set) and $n$ greater than the first proof of contradiction from the ZFC axioms.  The standard part of $B$ is clearly empty in every model of ZFC and $B$ clearly computable in any model of $ZFC+CON(ZFC)$ and not in any model of $ZFC+\lnot CON(ZFC)$.
