I'm reading through $C^{*}$ algebras by Murphy, and the following theorem is presented.
Let $I \subset A$ be a closed ideal of a $C^{*}$ algebra $A$. Then there exists a unique $*$ homomorphism $\varphi: A \longrightarrow M(I)$ extending the inclusion $I \rightarrow M(I)$, where $M(I)$ is the multiplier algebra of $I$. Moreover if $I$ is essential in $A$, then $\varphi$ is injective.
Then they state 'From this result we conclude that $M(I)$ is the "largest" unital $C^{*}$ algebra containing $I$ as a closed essential ideal.'
Why is this true? Im assuming what they mean is given any unital $C^{*}$ algebra $B \supset M(I)$, $I$ cannot be essential in $B$ if the inclusion $M(I) \subset B$ is proper, but I don't see how the above result tells us that.
It's literally the statement of the theorem. The theorem says that if $I\subset A$ as an essential ideal, then the inclusion $I\to M(I)$ extends to an injective inclusion $A\to M(I)$. So $I\subset\varphi(A)\subset M(I)$. That is, any C$^*$-algebra that contains $I$ as an essential ideal is naturally contained in $M(I)$.
When $B\supsetneq M(I)$, then $I$ (if represented inside $M(I)$) cannot be an ideal of $B$. This has nothing to do with the theorem, it's just the fact that if $B$ contains $b\not\in M(I)$, then by definition $bI\not\subset I$.