It depends on the function $f : i \mapsto |A_i|$. The ultraproduct $\prod A_i/ U$ has the same cardinality as $\prod f(i)/U$. $f$ represents an ordinal $\alpha <j(\kappa)$ in the ultrapower $M$ of the universe $V$, where $j : V \to M$ is the cnonical embedding. It's a standard fact that $2^\kappa < j(\kappa) < (2^\kappa)^+$. So the cardinality of $\prod f(i)/U$ is $|\alpha|$, which can be anything between $\kappa$ and $2^\kappa$, because we can select an $f$ corresponding to each such $\alpha$. For example, if $f$ is the identity function, then the ultraproduct has size $\kappa$. If $f(i) = i^+$ for $U$-almost all $i$, then $\prod f(i) / U = \kappa^+$.

It depends on the function $f : i \mapsto |A_i|$. The ultraproduct $\prod A_i/ U$ has the same cardinality as $\prod f(i)/U$. $f$ represents an ordinal $\alpha <j(\kappa)$ in the ultrapower $M$ of the universe $V$, where $j : V \to M$ is the canonical embedding. It's a standard fact that $2^\kappa < j(\kappa) < (2^\kappa)^+$. So the cardinality of $\prod f(i)/U$ is $|\alpha|$, which can be anything between $\kappa$ and $2^\kappa$, because we can select an $f$ corresponding to each such $\alpha$. For example, if $f$ is the identity function, then the ultraproduct has size $\kappa$. If $f(i) = i^+$ for $U$-almost all $i$, then $\prod f(i) / U = \kappa^+$.