Is anything known about the consistency strength of the statement:

"There is a normal measure (on a cardinal) that is not ordinal-definable"?

In particular, is it consistent relative to the existence of a measurable cardinal? It looks like it's consistent relative to the existence of a supercompact cardinal. If $\kappa$ is supercompact then we can force to make it Laver indestructible.

So assume that $\kappa$ is still $(\kappa+2)$-strong after we add $(2^{2^\kappa})^+$ many Cohen subsets of $\kappa^+$, more than the number of measures on $\kappa$ in $V$. Solovay proved that if $\kappa$ is $(\kappa+2)$-strong then for every set $X \in V_{\kappa+2}$ there is a normal measure on $\kappa$ whose ultrapower contains $X$. So letting $X$ range over the Cohen subsets of $\kappa^+$ that we added, a counting argument shows that we must get some normal measures on $\kappa$ that are not in $V$. Cohen forcing is homogeneous, so these measures cannot be ordinal-definable. I don't know how strong this kind of indestructibility is, or whether it's necessary.

I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.