 Consistency strength of a worldly cardinal

Let $T=\mathrm{ZFC}+\text{“there is a worldly cardinal”}$ and $T^{\ast }=\mathrm{ZFC}+\mathrm{Con}(\mathrm{ZFC})$. A worldly $\kappa$ makes $V_{\kappa }$ a model of ZFC, and arithmetic absoluteness for a rank-initial model makes it a model of $T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$. If $T^{\ast }$ proved $\mathrm{Con}(T)$, then the stronger theory $T$ would prove its own consistency, contradicting the Gödel second incompleteness theorem. Thus, assuming consistency, $T^{\ast }{<}_{\mathrm{Cons}}T$.