The unique matrix that satisfies $EA = B$ is the matrix that "swaps" the
first and third rows. It is given as
$$E=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0
\\ \end{bmatrix}.$$
Edit:
Due to a question in the comments, here comes a bit longer explanation.
(1) The rows of matrix $B$ and $A$ are the same, except for the fact that we have to swap
the first and the third row.
(2) $E$, defined above, is the special matrix that swaps the first and third rows of any $3 \times 3$ matrix $O$ when multiplied by it from the left. This can, for example, be seen by simple matrix multiplication
$$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0
\\ \end{bmatrix}\begin{bmatrix} p_1 & p_2 & p_3 \\ q_1 & q_2 & q_3 \\ r_1 & r_2 & r_3
\\ \end{bmatrix} = \begin{bmatrix} r_1 & r_2 & r_3 \\ q_1 & q_2 & q_3 \\ p_1 & p_2 & p_3
\\ \end{bmatrix}. $$
Hence, as a particular case, we also have $EA=B$. (Moreover, since $A$ and $B$ are non-singular matrices, the solution to the matrix equation $XA=B$ is unique: $X=BA^{-1}$,
calculating this we would again get $X=E$.)
(3) Elementary matrices (see definition here) differs from the identity matrix by one single elementary row operation. After swapping the first and third row of $E$ (which is an elementary row operation) we arrive to matrix
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\\ \end{bmatrix},$$
which is exactly the identity matrix. Hence $E$ is an elementary matrix.
Best Answer
It took me a good 20 minutes to type this, so I'm gonna be pissed af if you don't read it.
Take the matrix $\begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix}$ and add $2/3$ times the first row to the second. You get $\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let $E_1$ be the elementary row matrix corresponding to the row operation you just did:
$$E_1 = \begin{pmatrix}1 & 0 \\ \frac{2}{3} & 1 \end{pmatrix}$$
Notice that $E_1 \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$.
Next, take the matrix $\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$ and add $-\frac{3}{8}$ times the second row to the first. You get $\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let $E_2$ be the the elementary row matrix corresponding to the row operation you just did:
$$E_2 = \begin{pmatrix} 1 & -\frac{3}{8} \\ 0 & 1 \end{pmatrix} $$
Notice that $E_2 \begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$.
Next, take $\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$ and multiply the first row by $-\frac{1}{3}$. You get $\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let
$$E_3 = \begin{pmatrix} -\frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}$$
Notice $E_3 \begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3}\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$
Now take $\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix}$ and multiply the bottom row by $-\frac{3}{8}$. You get $I =\begin{pmatrix} 1 &0\\0 & 1 \end{pmatrix}$. Let
$$E_4 = \begin{pmatrix} 1 & 0 \\ 0 & -\frac{3}{8} \end{pmatrix}$$
And notice $E_4\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix} = I$.
Now
$$I = E_4\begin{pmatrix} 1 & 0 \\ 0 & \frac{8}{3} \end{pmatrix} = E_4 E_3\begin{pmatrix} -3 & 0 \\ 0 & \frac{8}{3}\end{pmatrix} = E_4E_3E_2\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$$
$$ = E_4E_3E_2E_1 \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix}$$
and so
$$ \begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix} = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}$$
where $E_1^{-1} = \begin{pmatrix} 1 & 0 \\ -\frac{2}{3} & 1 \end{pmatrix}, E_2^{-1} = \begin{pmatrix} 1 & \frac{3}{8} \\ 0 & 1 \end{pmatrix}, E_3^{-1} = \begin{pmatrix} -3 & 0 \\ 0 & 1 \end{pmatrix}, E_4^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & -\frac{8}{3} \end{pmatrix}$. Those are the elementary matrices you want.