= Solution
Write
$$
T=\mathrm{ZFC}+\text{“there is a worldly cardinal”},
\qquad
T^*=\mathrm{ZFC}+\operatorname{Con}(\mathrm{ZFC}).
$$
If $\kappa$ is a <worldly cardinal>, then $V_\kappa\models\mathrm{ZFC}$. The existence of this set model proves $\operatorname{Con}(\mathrm{ZFC})$ in the universe. By <arithmetic absoluteness for a rank-initial model>, the same formal consistency statement holds in $V_\kappa$. Hence $V_\kappa\models T^*$, so $T$ proves $\operatorname{Con}(T^*)$.
Conversely, suppose $T^*$ proved $\operatorname{Con}(T)$. The theory $T$ proves every axiom of $T^*$, since a worldly cardinal proves $\operatorname{Con}(\mathrm{ZFC})$. It would therefore also prove $\operatorname{Con}(T)$, contrary to the <Gödel second incompleteness theorem> when $T$ is consistent. Thus $T^*$ cannot prove $\operatorname{Con}(T)$, and
$$
\boxed{T^*<_{\mathrm{Cons}}T.}
$$
This is the <consistency strength of a worldly cardinal> comparison.
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