= Solution
{c}
Use the <quotient filtration> induced from the <order filtration of a Weyl algebra>. The key step is to identify the entire <associated graded module> of the <left ideal> $I$, rather than merely take symbols of its displayed generators.
For every member $x^\alpha y^\beta$ of the <ordered monomial basis of a Weyl algebra>, right multiplication by the pure $y$-<monomials> gives
$$
(x^\alpha y^\beta)y_1^3y_2^2=x^\alpha y^{\beta+(3,2,0,\ldots,0)},\qquad
(x^\alpha y^\beta)y_1^2y_3^3=x^\alpha y^{\beta+(2,0,3,0,\ldots,0)}.
$$
No commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex <linear span> of the <basis> <monomials> satisfying either
$$
\beta_1\geq3,\ \beta_2\geq2,
\qquad\text{or}\qquad
\beta_1\geq2,\ \beta_3\geq3.
$$
Intersecting this <linear span> with each filtration piece and then taking symbols therefore yields the <monomial ideal>
$$
\operatorname{gr}I=K=(\xi_1^3\xi_2^2,\xi_1^2\xi_3^3)\subseteq A.
$$
By <3/b>,
$$
\operatorname{gr}(R/I)\cong A/K.
$$
To determine its <Krull dimension>, compute the <radical of an ideal>:
$$
\sqrt K=(\xi_1\xi_2,\xi_1\xi_3)
=(\xi_1)\cap(\xi_2,\xi_3).
$$
Indeed, $K$ lies in the displayed intersection, an intersection of <prime ideals>. Conversely, $(\xi_1\xi_2)^3\in K$ and $(\xi_1\xi_3)^3\in K$, so both generators of that intersection lie in $\sqrt K$. The equality of the intersection with $(\xi_1\xi_2,\xi_1\xi_3)$ follows by factoring out $\xi_1$ and reducing modulo $(\xi_2,\xi_3)$.
The <order characteristic variety of a Weyl algebra module> in this case has two <irreducible components>:
$$
V(K)=\{\xi_1=0\}\ \cup\ \{\xi_2=\xi_3=0\}.
$$
They are coordinate <affine spaces> of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\ldots,x_n,\xi_1,\ldots,\xi_n$. The <Krull dimension> is the larger of these component dimensions. Thus
$$
\boxed{d(R/I)=2n-1.}
$$
In particular, two displayed generators do not force a drop of two in <dimension>: their common factor creates the component $\xi_1=0$ of dimension $2n-1$.
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