= Solution
{c}
Route a directed <ghost loop> so that its internal <four-momenta> are $l$ and $l+k$. At the external <gluon> with indices $(\mu,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other <Feynman vertex> has the reverse order. The stated ghost-gluon <Feynman rules> supply $g_D^2f^{acd}f^{bdc}(l+k)^\mu l^\nu$. Their <colour factor> is
$$
f^{acd}f^{bdc}=-C_A\delta^{ab}.
$$
The two <ghost propagators> supply $i^2=-1$, while the closed <ghost loop> supplies another $-1$. Thus, with the insertion convention stated above,
$$
i\Pi_{\mathrm{gh}}^{\mu\nu,ab}=-g_D^2C_A\delta^{ab}I^{\mu\nu},\qquad
I^{\mu\nu}=\int\frac{d^Dl}{(2\pi)^D}\frac{(l+k)^\mu l^\nu}{(l^2+i0)((l+k)^2+i0)}.
$$
This establishes every prefactor before integration. To isolate its <ultraviolet pole>, take generic <off shell> $k$, initially $k^2<0$, so the massless integral has no overlapping <infrared divergence>. Introduce a <Feynman parameter>, shift $r=l+xk$, and set $\Delta=-x(1-x)k^2>0$. Odd powers of $r$ vanish, giving
$$
I^{\mu\nu}=\int_0^1dx\left[\frac{g^{\mu\nu}}{D}\int\frac{d^Dr}{(2\pi)^D}\frac{r^2}{(r^2-\Delta+i0)^2}-x(1-x)k^\mu k^\nu\int\frac{d^Dr}{(2\pi)^D}\frac1{(r^2-\Delta+i0)^2}\right].
$$
The needed <dimensional regularization> integrals follow from the supplied <gamma function> formula. In $D=4-\epsilon$,
$$
\begin{aligned}
J_1(\Delta)&=\int\frac{d^Dr}{(2\pi)^D}\frac1{r^2-\Delta+i0}
=\frac{i\Delta}{16\pi^2}\frac2\epsilon+O(1),\\
J_2(\Delta)&=\int\frac{d^Dr}{(2\pi)^D}\frac1{(r^2-\Delta+i0)^2}
=\frac{i}{16\pi^2}\frac2\epsilon+O(1),\\
\int\frac{d^Dr}{(2\pi)^D}\frac{r^2}{(r^2-\Delta+i0)^2}&=J_1+\Delta J_2=\frac{i}{16\pi^2}\frac{4\Delta}{\epsilon}+O(1).
\end{aligned}
$$
Here $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, rather than $1/\epsilon$. Using $1/D=1/4+O(\epsilon)$ and $\int_0^1x(1-x)\,dx=1/6$, obtain
$$
I^{\mu\nu}\big|_{\mathrm{UV}}=-\frac{i}{96\pi^2\epsilon}\left(k^2g^{\mu\nu}+2k^\mu k^\nu\right).
$$
Therefore the <ghost-loop pole of the gluon self-energy> is
$$
\boxed{\Pi_{\mathrm{gh}}^{\mu\nu,ab}\big|_{\mathrm{UV}}=\frac{g^2C_A\delta^{ab}}{96\pi^2\epsilon}\left(k^2g^{\mu\nu}+2k^\mu k^\nu\right).}
$$
The factor $\mu^\epsilon$ in $g_D^2$ affects only finite terms at this order. The isolated <ghost loop> is not transverse: contracting this result gives $g^2C_A\delta^{ab}k^2k^\nu/(32\pi^2\epsilon)$. Transversality is a condition on the complete <gluon self-energy>, including the other loop graphs.
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