It is known that the property of being a worldly cardinal is not absolute (a cardinal $\kappa$ is worldly iff $V_{\kappa} \vDash \textsf{ZFC}$). See here and here for more.

This said, it is the case that the existence of two worldly cardinals has strictly greater consistency strength than the existence of one worldly cardinal, and any proof I can see of this will require that if $\kappa_1 < \kappa_2$ are worldly cardinals, then $V_{\kappa_2}$ must believe $\kappa_1$ is worldly. (Easiest proof: show that $V_{\kappa_2}$ believes that $V_{\kappa_1} \vDash \textsf{ZFC}$.)

I wasn’t sure how to show that this was the case — I was thinking that maybe because $V_{\kappa_2}$ is transitive it must agree with $V$ on whether $V_{\kappa_1}$ models $\textsf{ZFC}$, but I’m not totally clear on/sure of this.