= Solution
Give $U(\mathfrak g)$ its degree <filtration>. The <Poincaré-Birkhoff-Witt theorem> identifies its <associated graded ring> with $\operatorname{Sym}_k\mathfrak g$, 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> $\operatorname{gr}I$ is a homogeneous <right ideal> of $\operatorname{gr}A$. Choose finite homogeneous generators and lift them to $u_1,\ldots,u_m\in I$. 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 $\sum_i u_ia_i$. The remainder has degree strictly below $d$. Induction, terminating because degrees are nonnegative, proves $I=\sum_i u_iA$. 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 \b[$R$ 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.
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