Given a C*-algebra $\mathcal{A}$.
Multiplier Algebra: $$\mathcal{M}(\mathcal{A}):=\{(L,R)\in\mathcal{B}(\mathcal{A})^2:\ldots\}$$
We know already:
It is the largest unitization of the algebra.
A unitization has the algebra as essential ideal.
For unital algebras: $$1\in\mathcal{A}:\quad\mathcal{A}=\mathcal{M}(\mathcal{A})$$
Clearly we have:
Then the smallest unitization is the algebra.
How to check that it is also the multiplier algebra?
$\mathcal A$ is an ideal in $\mathcal{M(A)}$. Note that the double centraliser corresponding to $1$ is the unit in $\mathcal{M(A)}$, so $\mathcal A$ is an ideal that contains the unit and so must be the entire space.
To see that with: $L_1(a)=1\cdot a$, $R_1(a)=a\cdot 1$ one gets that $L,R\ =\mathrm{id}$ and then $(L,R)=1_{\mathcal{M(A)}}$.