Let $(R,m,k)$ be a local Gorenstein ring of dimension 0, that is $R$ has finite injective dimension (see for instance the definition in Cohen-Macaulay Rings from Bruns and Herzog).
In Definition 3.3.1 in Cohen-Macaulay Rings from Bruns and Herzog a canonical module of $R$ is defined as a maximal Cohen-Macaulay module of type 1 and of finite injective dimension.
Moreover, in Theorem 3.3.7 in the same book we have
If $(R,m,k)$ is Cohen-Macaulay, then $R$ is Gorenstein if and only if a canonical module $\omega_R$ exists and is isomorphic to $R$, i.e. $\omega_R \cong R$.
Now in Eisenbuds book Commutative Algebra with a View Towards Algebraic Geometry the author defines
a zero-dimensional local ring $(R,m,k)$ is Gorenstein if and only if the injective hull $E(k)$ of $k$ is isomorphic to $R$.
Now since I assume that these definitions should be consistent, there should be some reference that proves that for local Gorenstein rings $(R,m,k)$ of dimension 0 (using the definition of Bruns and Herzog) the injective hull $E(k)$ is isomorphic to $R$.
Best Answer
Over an arbitrary CM ring you have $\Omega$ a canonical module if $$\text{Ext}_{R}^{n}(k,\Omega)\simeq \begin{cases} k &\mbox{ if } n=\text{dim}\,R;\\ 0 &\mbox{ otherwise } \end{cases} $$ Now, if $\text{dim}\,R=0$ and $\Omega$ is canonical then $\Omega$ is injective by the definition and the fact there are no other prime ideals. Yet then $\Omega\simeq E(k)^{(n)}$ for $n=\text{dim Hom}_{R}(k,\Omega)$, but this is just equal to 1 by definition so $\Omega\simeq E(k)$.
But if $R$ Gorenstein then $R\simeq \Omega$ which is then isomorphic to $E(k)$ if $\text{dim}\,R=0$.
Edit:
Over any commutative noetherian ring $R$ and for any $R$-module $M$, it is known that $$E(M)\simeq\bigoplus_{\mathfrak{p}\in\text{Spec}R}E(R/\mathfrak{p})^{(\mu_{\mathfrak{p}})}$$ where $\mu_{\mathfrak{p}}=\text{dim}_{R_{\mathfrak{p}}}\text{Hom}_{R}(R/\mathfrak{p},M)_{\mathfrak{p}}$. In our case (when $R$ is local of dimension zero), we know $\Omega$ is injective so is its own injective hull, and $\mu_{\mathfrak{m}}= \text{dim Hom}_{R}(k,\Omega)=1 $ as $R_{\mathfrak{m}}=R$.