is the existence of an inaccessible cardinal stronger than just CON(ZFC)? [closed]

is it even stronger than that ZFC has a transtitive model?

• What have you done to research these questions, by the way? Following the several links that you (here and on MSE) were given to Cantor's Attic; or searching on Google would have revealed several answers, I am sure. – Asaf Karagila Mar 25 '14 at 9:24
• Also, what is your motivation for asking this question (it completely lacks any context, or as I pointed out, research effort)? Is it idle curiosity, or did it come up somewhere? – Asaf Karagila Mar 25 '14 at 9:31

If $\kappa$ is a Worldly cardinal, then $V_\kappa$ is a model of $\sf ZFC$, indeed a transitive model. And therefore $\rm Con\sf (ZFC)$ holds. As you already know, the least Worldly cardinal is not inaccessible, so if $\kappa$ is the least Worldly, there is no inaccessible cardinal in $V_\kappa$.
Take a countable elementary submodel of such $V_\kappa$, and collapse it, then you have a countable transitive model and its height is some countable ordinal $\alpha$. Consider the $\alpha_0$ to be the least height of a transitive model, and $\alpha_1$ the second-least. Then $L_{\alpha_0}\in L_{\alpha_1}$, and therefore $L_{\alpha_1}$ is a model of $\sf ZFC$ which knows that there is a transitive model in the universe.