Suppose we have a first order theory $\text{T}$ such that
Does that mean that $\text{T}$ must prove existence of a large cardinal?
If Yes, then among the list of the known large cardinal proeprties , which one is the least that $\text{T}$ must prove its existence?
On
Let me interpret your question in a slightly different way, which may be what you thought about.
$T$ extends $\sf ZF$, and if $M$ is a model of $T$, then there is no model extending it with the same ordinals which satisfies $V=\rm HOD$.
In this sense it's a sensible question, since every model of $\sf ZFC$ extends to a model of $\rm HOD$.
The answer is no, though. Simply because there are models of $\sf ZF$ that cannot be extended to models of $\sf ZFC$ without adding ordinals. The obvious example would be Gitik's model where all the ordinals have cofinality $\omega$. But of course that we can prove there are very large cardinals in some inner models. So it doesn't help.
However, D. B. Morris proved in his thesis, and later I rediscovered a similar proof, that it is consistent that there is an extension of every model of $\sf ZFC$ to a model where for every infinite ordinal $\alpha$, there is a set $A_\alpha$ which is a countable union of countable sets, and the power set of $A_\alpha$ can be mapped onto $\alpha$. It follows that this model cannot be extended to a model of $\sf ZFC$ without adding ordinals.
In particular, it cannot be extended to a model of $V=\rm HOD$.
This is just asking if $V\ne HOD$ implies any large cardinal properties. If we start with a model with no inaccessibles and then force $V\ne HOD$ (by, say, adding a Cohen real), the resulting outer model has no inaccessibles either (inaccessibility is absolute that way). Thus $V\ne HOD$ does not imply the existence of inaccessibles.