Let $\mathsf{ZFC}^-$ be the Zermelo-Fraenkel set theory without power set axiom.
For a transitive model $M$ of $\mathsf{ZFC}^-$ and an cardinal $\kappa\in M$ in the sense of $M$, an unary predicate $U$ over $M$ is an $\kappa$-complete ultrafilter if $M$ thinks it is $\kappa$-complete and $U$ is weakly amenable, that is, for any $F\in {^\kappa}M\cap M$, $\{\xi<\kappa : F(\xi)\in U\}\in M$. 


$M$ thinks $\kappa$ is weakly compect. I think in fact we can show $M$ thinks $\kappa$ is Ramsey. However it does not tell about their consistency strength as $\mathsf{ZFC}^-$ is far more weaker than $\mathsf{ZFC}$.


My question is, the exact consistency strength of the existence of the model $M\models \mathsf{ZFC}^-$ with a $\kappa$-complete ultrafilter over $M$ is known? Theorem 1.1 [of an article by Gitman](https://arxiv.org/pdf/0801.4723.pdf) and close relation between $\kappa$-complete ultrafilter and elementray embedding suggests its consistency strength may be related to the existence of a weakly compact cardinal.



Consistency strength of the existence of a transitive model of $\mathsf{ZFC}^-$ with a $\kappa$-complete ultrafilter
Let's make some additional assumptions on the ultrafilter $U$. Suppose $M\models{\rm ZFC}^-$ and $\kappa$ is a cardinal in $M$. We say that $U$ is an $M$-*ultrafilter* if $\langle M,\in,U\rangle$ satisfies that $U$ is a $\kappa$-complete normal ultrafilter on $\kappa$. Because $U$ is only $\kappa$-complete for sequences in $M$ and $M$ might be missing even countable sequences, the ultrapower of $M$ by $U$ may not be well-founded. So let's say that $U$ is *good* if the ultrapower by $U$ is well-founded.


The consistency of the existence of $M\models{\rm ZFC}^-$ for which there is a good weakly amenable $M$-ultrafilter $U$ on a cardinal $\kappa$ in $M$ is between an ineffable cardinal and an $\omega$-Erdős cardinal, which is much weaker than a Ramsey cardinal. In particular, this assumption is consistent with $V=L$. 


Here is a sketch of the argument. Suppose $j:M\to N$ is the ultrapower by $U$ (with $N$ transitive by assumption) with critical point $\kappa$ (by normality). By weak amenability, in $N$, $\kappa$ has the property that every collection of $\kappa$-many subsets of $\kappa$ has a $\kappa$-complete filter. Thus, by elementarity $M$ satisfies that this property holds for cofinally many $\alpha<\kappa$. Now, we argue that each such $\alpha$ is weakly compact in $L\_\kappa\models{\rm ZFC}$. Next, we can argue that $\kappa$ has the ineffability property in $N$, and therefore there are cofinally many $\alpha$ that are ineffable in $L\_\kappa$. The reason I am going down to $L$ is that we did not assume that $M$ has any powersets and therefore even the notion of inaccessible cardinal does not make sense for $M$. For the upper bound, an $\omega$-Erdős cardinal implies the consistency of what I call a $1$-*iterable* cardinal $\kappa$ having the property that every $A\subseteq\kappa$ is contained in a model $M\models{\rm ZFC}^-$ of size $\kappa$ with $\kappa\in M$ for which there is a good weakly amenable $M$-ultrafilter on $\kappa$.


If we don't assume that $U$ is normal or that the ultrapower by $U$ has to be well-founded, the consistency strength goes down significantly. At most a weakly compact suffices. If $\kappa$ is weakly compact and $M\_0\models{\rm ZFC}^-$ has size $\kappa$, then we can find a $\kappa$-complete ultrafilter $U\_0$ for $M\_0$ with a well-founded ultrapower. Moreover, if $M\_1\models{\rm ZFC}^-$ has size $\kappa$ and extends $M\_0$, then we can extend $U$ to a $\kappa$-complete ultrafilter $U\_1$ for $M\_1$. Thus, in $\omega$-many steps, we can obtain $M=\bigcup\_{n\in\omega}M\_n$ and a weakly amenable $U$ that is $\kappa$-complete for sequences from $M$. The argument for the "moreover" part can be found [here](http://www.math.uni-bonn.de/people/pholy/ramseylike.pdf) in an article of Peter Holy and Philipp Schlicht. I suspect that you don't even need a weakly compact. (I am assuming $\kappa$ uncountable here). 


