= Solution
Choose a generator $u$ of the <cyclic group> $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital <ring homomorphism> $\mathbb F_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a <commutative ring>, evaluating a <polynomial> at $u$ defines a <ring homomorphism>
$$
\phi:\mathbb F_2[X]\longrightarrow R,\qquad\sum_ja_jX^j\longmapsto\sum_ja_ju^j.
$$
Because $\phi(X^p-1)=0$, it factors through the <quotient ring> to give
$$
\boxed{\overline\phi:\mathbb F_2[X]/(X^p-1)\longrightarrow R,\qquad\overline X\longmapsto u}.
$$
Its image is the <subring> $S=\mathbb F_2[u]$. Every <unit> of $S$ is a <unit> of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. Hence
$$
\boxed{U(S)=U(R)}.
$$
The inverse argument matters: merely belonging to a <subring> would not by itself make an ambient <unit> invertible there.
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