If I have a multiplication operator $L_A: \mathbb{R}^4 \to \mathbb{R}^4$, where $$ A=\begin{pmatrix} 43 & -32 & 35 & 14 \\ -66 & 44 & -52 & -18 \\ -137 & 97 & -110 & -41 \\ 59 & -43 & 48 & 18\end{pmatrix} $$
how do I find a matrix $B$ whose row space is smallest $L_A$-invariant subspace that contains the vector $(0,2,1,2)$. Do I need to find the eigenvalues and the eigenvectors?
I assume that the map you have in mind is $L_A(x) = Ax$. If this is not what you have in mind, please edit you question to clarify what you do mean.
If $v$ is the column vector $v = (0,2,1,2)$, then the matrix $$ M = \pmatrix{v & Av & A^2 v & A^3 v} $$ has column span equal to the smallest $L_A$-invariant subspace containing $v$. Thus, $B = M^\top$ has the desired properties.
When following the above process, I end up with the answer $$ B = \left(\begin{matrix}0 & -1 & -1 & 0\\2 & 0 & -2 & 12\\1 & 2 & -1 & 12\\2 & -2 & 1 & -3\end{matrix}\right). $$ Unlike the answer that you found, this matrix has rank 3.
You can also row-reduce this matrix to find another possibility, namely $$ \left(\begin{matrix}1 & 0 & 0 & 3\\0 & 1 & 0 & 3\\0 & 0 & 1 & -3\end{matrix}\right). $$