Warning: the following is just what I found from some work on MathSciNet, following Yemon's hint in the comments. It's not meant to be accurate historical notes.

A "dual" $C^\ast$-algebra is defined as follows. Let $A$ be an algebra and for a subset $M\subseteq A$ let $R(M) = \{ x\in A : Mx=\{0\}\}$; similarly define $L(M) = \{ x\in A : xM=\{0\}\}$. Then a $C^\ast$-algebra is *dual* if for each closed left ideal $I$ we have that $L(R(I))=I$. (The involution can be used to show that the analogous definition with right ideals gives the same notion).

An early paper which studied these is Kaplansky, The structure of certain operator algebras, see section 2. It seems that Berglund, Ideal $C^\ast$-algebras was the first to obtain the equivalence you seek (why is the Duke journal archive behind a paywall??) A short proof is in McCharen, A characterization of dual $B^{\ast}$-algebras These give the following:

**Claim:** A $C^\ast$-algebra $A$ is an ideal in its bidual if and only if $A$ is dual.

We now combine this with known characterisations of dual $C^\ast$-algebras:

**Claim:** A $C^\ast$-algebra $A$ is dual if and only if $A$ is isomorphic to a $C^\ast$-subalgebra of $K(H)$ for some $H$ if and only if $A$ is the $c_0$-direct sum of algebras of the form $K(H)$ for some (finite or infinite dimensional) $H$.

Thus $K(H)$ and $c_0$ really are the archetypal examples. These results are quoted in the paper of Kaplansky I linked above, and in Dixmier's book (English edition, section 4.7.20) though in both cases it seems only further references are given, not proofs. I am afraid that I don't know of a modern, self-contained treatment.

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