 Solution

Give $U(\mathfrak{g})$ its degree filtration. The Poincaré-Birkhoff-Witt theorem identifies its associated graded ring with ${\mathrm{Sym}}_k\mathfrak{g}$, a polynomial ring in ${\dim }_k\mathfrak{g}$ variables. The Hilbert basis theorem makes this graded algebra Noetherian on both sides.

Here is the required transfer, including why a finite list of symbols gives actual generators. For a right ideal $I$ of a nonnegatively filtered ring $A$, its induced graded ideal $\mathrm{gr}I$ is a homogeneous right ideal of $\mathrm{gr}A$. Choose finite homogeneous generators and lift them to $u_1,\dots ,u_m\in I$. For $u\in I$ of degree $d$, express its leading symbol as a graded right linear combination of these symbols, using homogeneous coefficients of complementary degrees. Lift the coefficients to $A$ and subtract the corresponding ${\sum }_i{u}_i{a}_i$. The remainder has degree strictly below $d$. Induction, terminating because degrees are nonnegative, proves $I={\sum }_i{u}_{i}A$. The identical argument with coefficients on the left treats left ideals.

This ascending filtered-graded transfer of Noetherianity proves that $U(\mathfrak{g})$ is right and left Noetherian. A quotient ring inherits either property: pull an ascending ideal chain back to the original ring, or take images of ideal generators. Part (a) therefore gives $R$ right and left Noetherian. Alternatively, under the broader almost commutative algebra convention, its commutative finitely generated associated graded ring is directly a quotient of a finite-variable polynomial ring, and the same leading-symbol argument proves the conclusion.