 Solution

Let $M$ be an externally countable model of ZFC, and let $\kappa =({\omega }_1{)}^M$. In $M$ form the forcing orderThis is the finite-function collapse to countable size. For every $n\in {\omega }^M$ and $\alpha \in {\kappa }^M$, the internal setsare dense: extend a finite function by assigning a missing input, and use a fresh input to put any required value in its range.

Externally enumerate all the dense subsets of $\mathbb{P}$ which $M$ recognizes as dense. There are only countably many, because $M$ is countable. Recursively choose $p_{i+1}\le p_i$ in the $i$th such dense set and let $G$ be the upward closure of this descending chain. This produces a generic filter over $M$. By the forcing theorem, its generic extension $M[G]$ satisfies ZFC. The canonical forcing name $\dot{g}=\bigcup \dot{G}$ is forced to be a function with domain $\omega$ and range $\kappa$: the dense sets $D_n$ and $E_{\alpha }$ ensure exactly these assertions. ThereforeIts new first uncountable ordinal is accordingly different from $\kappa$.

For a transitive $M$, ordinary evaluation of names gives the familiar literal extension. The question only assumes a countable model, so one must also cover externally ill-founded models. In that case use the quotient of the internally defined forcing names by forced equality modulo $G$, with membership defined by forced membership. The internal forcing identities and the forcing theorem hold in $M$ and give a well-defined first-order structure satisfying every standard ZFC axiom. Check names embed the original membership structure, so identify their images with $M$. Its domain is a quotient of a subset of the countable domain of $M$, hence is externally countable. Thus the required extension exists even without assuming transitivity. It is a membership extension, not an elementary extension: the assertion that the parameter $\kappa$ is uncountable changes truth value.