= Solution
{c}
\b[Yes.] Taking the constant coefficient defines a surjective <ring homomorphism>
$$
\epsilon:R[[x]]\longrightarrow R,\qquad
\sum_{n\geq0}a_nx^n\longmapsto a_0.
$$
Its <kernel of a ring homomorphism> is $(x)$, so the <first isomorphism theorem for rings> gives $R[[x]]/(x)\cong R$.
We prove the needed preservation result. If $A$ is a <Noetherian ring> and $J\subseteq A$ is an <ideal>, any ascending chain of <ideals> in $A/J$ lifts under the quotient <ring homomorphism> to an ascending chain of <ideals> in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every <quotient ring> of a <Noetherian ring> is <Noetherian>.
Applying this with $A=R[[x]]$ and $J=(x)$ gives
$$
\boxed{R[[x]]\text{ Noetherian}\ \Longrightarrow\ R\text{ Noetherian}.}
$$
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