Your procedure seems correct, but it depends on how you write down these matrices.
Let $(e_i)_{i=1,2,3}$ and $(e'_j)_{j=1,2}$ denote the canonical bases in the source and the image. Let $B=(u_\ell)_{\ell=1,2,3}$ and $B'=(u'_k)_{k=1,2}$ denote the stated bases vectors and finally $A=(a_{kl})$ the matrix calculated in the bases $B,B'$. (As a paranthetical remark: One of my predelictions concerning change of base calculations is to write an abstract vector $x$ in the form $x=\sum_k u_k x_k$ in the basis $B$, i.e. with the vectors to the right. This ensures keeping correct track of indices and matrix multiplications. Paranthesis closed).
The abstract linear transformation acts as follows:
$$ f(u_\ell) = \sum_{k=1}^2 u'_k A_{k\ell} $$
Writing $u_\ell= \sum_{i=1}^3 e_i u_{i\ell}$ and $u'_k= \sum_{j=1}^2 e'_j u'_{jk}$ we get the identity:
$$ \sum_i f(e_i) u_{i\ell} = \sum_j e'_j \sum_k u'_{jk} A_{k\ell} $$
Let $M$ be the matrix of $f$ between the canonical bases. Then
$$ \sum_i f(e_i) = \sum_j e'_j M_{ji}$$
So comparing we get: $\sum_i M_{ji} u_{i\ell} = \sum_k u'_{jk} A_{k\ell}$ or
with your data:
$$ M
\left[\begin{matrix} 1 & 0 & 1\\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} \right]=
\left[\begin{matrix} 1 & 0\\ 1 & 1 \end{matrix} \right]
\left[\begin{matrix} 2 & 1 & 3\\ 3 & 1 & -3 \end{matrix} \right]
$$
which you then have to solve for $M$. The answer is then simply $M'$ the usual transpose of $M$.
Using scilab I got:$$ M=
\left[\begin{matrix} 2 & 1 & -1\\ -2 & 2 & 5 \end{matrix} \right]$$
so the answer should be its transpose. You may now tell me if this is compatible with the answer to the exercise?
