One entry of an inverse

You do not need the whole inverse: one cofactor and the determinant are enough, as long as you get the indices the right way round.

1. Expanding along row $1$ gives $\det (A)=1\times (1\times 1-3\times 0)-2\times (0\times 1-3\times 2)+0=1+12=13$.
2. The entry in row $i$, column $j$ of $A^{-1}$ is the cofactor $C_{ji}$ divided by $\det (A)$, with the indices swapped.
3. So we need $C_{12}$: deleting row $1$ and column $2$ leaves $\left[\begin{array}{cc}0& 3\\ 2& 1\end{array}\right]$, whose determinant is $0\times 1-3\times 2=-6$.
4. The sign at position $(1,2)$ is negative, so $C_{12}=6$ and the entry is $\frac{6}{13}$.

Answer: $\frac{6}{13}$