Answer
$x_{\text {destroyed }} =q_{\text {in }}\left(\eta_{\text {th, C }}-\eta_{\text {th }}\right)
$
Work Step by Step
The exergy destruction associated with a cycle is given on a unit mass basis as $$
x_{\text {destoyed }}=T_0 \sum \frac{q_R}{T_R}
$$ where the direction of $q_{\text {in }}$ is determined with respect to the reservoir (positive if to the reservoir and negative if from the reservoir). For a cycle that involves heat transfer only with a source at $T_{\mathrm{H}}$ and a sink at $T_0$, the irreversibility becomes $$
\begin{aligned}
x_{\text {destroyed }} & =T_0\left(\frac{q_{\text {out }}}{T_0}-\frac{q_{\text {in }}}{T_H}\right)=q_{\text {out }}-\frac{T_0}{T_H} q_{\text {in }}=q_{\text {in }}\left(\frac{q_{\text {out }}}{q_{\text {in }}}-\frac{T_0}{T_H}\right) \\
& =q_{\text {in }}\left[\left(1-\eta_{\text {th }}\right)-\left(1-\eta_{\text {th }, C}\right)\right]=q_{\text {in }}\left(\eta_{\text {th}, C}-\eta_{\text {th }}\right)
\end{aligned}
$$
