Show that $\mathbb Z[x] / I \cong \overline{\mathbb Z}[x]/ \overline{I}$, where $\overline{\mathbb Z} = \mathbb Z /3 \mathbb Z$ and $\overline{I}$ is the ideal generated by the polynomial $x^3 -x^2+2x-1$ over $\mathbb Z /3 \mathbb Z$.
Also, I would love to know how to conclude afterwards that $\mathbb Z[x] / I$ is a field, and will its order be $3$?
I've already shown that the ideal $I = 3\mathbb Z[x] + (x^3-x^2+2x-1)\mathbb Z[x]$ is not a principal ideal of $\mathbb Z[x]$, how can I continue from here?
As I commented in the linked question; indeed $\mathbb Z[x] / I \cong \overline{\mathbb Z}[x]/ \overline{I}$, as can be seen by the chain $$\Bbb{Z}[x]/(3,x^3-x^2+2x-1)\cong(\Bbb{Z}[x]/3\Bbb{Z}[x])/(x^3-x^2+2x-1)\cong(\Bbb{Z}/3\Bbb{Z})[x]/(x^3-x^2+2x-1).$$ Try to construct these isomorphisms explicitly if any of them is not clear.
The order of this quotient is $3^3=27$, because its elements are (represented by) polynomials of degree less than $3$ with coefficients in $\Bbb{Z}/3\Bbb{Z}$.
The quotient is a field if and only if the ideal $\overline{I}\subset\overline{\Bbb{Z}}$ is maximal, or equivalently, if and only if the polynomial $x^3-x^2+2x-1$ is irreducible over $\Bbb{Z}/3\Bbb{Z}$. A cubic polynomial is irreducible if and only if it has no roots, and a quick check shows that this polynomial indeed has no roots in $\Bbb{Z}/3\Bbb{Z}$.