= Solution
For a displacement <group representation>, an atom contributes only if the operation leaves its site fixed. Its three Cartesian displacements then contribute the trace of the polar-vector transformation.
Take $z$ perpendicular to the ring and an in-plane $x$ axis through opposite <carbon> atoms. To avoid confusing the principal twofold rotation with the in-plane axes, write $C_2(z)$ for the former, $C_{2a}$ for an in-plane axis through atoms and $C_{2b}$ for one through opposite bond midpoints. The <carbon> displacement characters in the <D6h molecular point group> are
$$
\begin{array}{c|rrrrrr}
C&E&2C_6&2C_3&C_2(z)&3C_{2a}&3C_{2b}\\\hline
\chi_C&18&0&0&0&-2&0\\
iC&i&2S_3&2S_6&\sigma_h&3\sigma_d&3\sigma_v\\\hline
\chi_C(iC)&0&0&0&6&0&2
\end{array}
$$
A through-atom half-turn fixes two <carbon> sites and has Cartesian trace $-1$, giving $-2$. The molecular-plane reflection fixes all six carbons with trace $+1$, giving $6$. A through-atom vertical mirror fixes two sites with trace $+1$, giving $2$.
Using <molecular character reduction>, the 18-dimensional carbon-only result is
$$
\boxed{\Gamma_C=A_{1g}+A_{2g}+B_{2g}+E_{1g}+2E_{2g}
+A_{2u}+B_{1u}+B_{2u}+2E_{1u}+E_{2u}.}
$$
The six one-dimensional terms contribute six dimensions and the six copies of two-dimensional representations contribute twelve, totaling eighteen. The distinction between carbon-only and whole-molecule displacements is needed in part (ii).
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