Suppose that ZFC is consistent, and let ZFC'=ZFC+Con(ZFC). We can define a reasonable notion of the length of a proof inside ZFC', such that for any $n$ the set $P_n$ of all proofs of length $\leq n$ is finite. For any provable statement $\phi$ (I mean provable in $ZFC'$), denote by $c(\phi)$ the shortest possible length of a proof of $\phi$.
If we denote by $L_n$ the (finite) set of all provable statements written with at most $n$ characters, then
the map $f_1$ defined by $f_1(n)={\sf max}_{\phi \in L_n} (c(\phi))$ is obviously not computable (otherwise provability in ZFC' would become decidable : we would only need to browse all proofs of length $\leq f_1(n)$).
On the other hand, the map $f_2$ defined by $f_2(n)={\sf max}_{\phi \in P_n} (c(\phi))$ is certainly computable (see fgp’s answer below).
Now consider a slightly more complicated case : denote by $Q_n$ the set of statements such that there is a proof (in ZFC') of $ZFC \vdash \phi$, whose length is $\leq n$. Is the map $f_3$ defined by $f_3(n)={\sf max}_{\phi \in Q_n} (c(\phi))$ computable ?
It certainly seems so. If you pick an $\phi \in P_n$, then by definition it has a proof of length $\leq n$. Thus, $f_2(n) \leq n$, so to compute $f_2(n)$ you only have to find the smallest $m \leq n$ such that $P_m$ proves the same statements as $P_n$.
In other words, if $\Phi_i$ is the (finite!) set of statements proven by $P_i$, then $f_2$ can be defined as $$ f_2(n) = \min \left\{ m \leq n\,:\, \Phi_m = \Phi_n \right\} \text{,} $$ which certainly is computable.