When is it $C(X)$? Suppose that $\tilde{X}$ is a compact space. If $C(\tilde{X})$ is isometrically isomorphic to the second dual of a Banach space, does there exist a locally compact space $X$ such that $C(\tilde{X})=C_0(X)^{**}$?
P.S. Note that the converse is always true, namely if $X$ is a locally compact space, there is a compact space $\tilde{X}$ such that $C_0(X)^{**}=C(\tilde{X})$. The Banach space $C_0(X)^{**}$ turns into a Banach algebra with the first (second) Arens product. This is a non-trivial product on $C_0(X)^{**}$.
 A: Only a partial answer, but too long for comments.
This was question 3 considered at a BIRS meeting in 2012 — I wonder if your question is motivated by seeing it here or raised elsewhere by Dales or his collaborators?
In any case, according to the final report (see the sidebar) the answer is yes when $C(\widetilde{X})$ is assumed to be isometric to the bidual of a separable Banach space, by old results of H. E. Lacey:
H. E. Lacey. A note concerning $A^{\ast} =L_1(\mu )$. Proc. Amer. Math. Soc. 29 (1971) 525—528.
Link to paper.
If I understand Lacey's result correctly, it says that if $C(\widetilde{X})=A^{**}$ for $A$ a separable Banach space then $C(\widetilde{X})$ is the bidual of $c_0$ or of $C[0,1]$.
It appears that at the time that report was written, the general case was still open. Perhaps if Fred Dashiell is still on MathOverflow he might be able to inform us of any further progress.
A: This is a detailed version of a previous comment that gives a partial answer. 
If $K$ is a compact space and $C(K)$ is isometrically isomorphic to $X^{**}$, then $X$ is a $\mathcal{L}_{\infty}$-space. Hence $X^{**}$ is injective. 
It was proved by R. Haydon [Israel J. Math. 31 (1978), 142-152] that an injective bidual space is isomorphic to $\ell_\infty(\Gamma)$ for some set $\Gamma$. Hence $C(K)$ is isomorphic to $c_0(\Gamma)^{**}$. 
I do not know if $C(K)$ is isometrically isomorphic to $c_0(\Gamma)^{**}$. 
