Does every standard set have a standard cardinality? Add a new primitive one place predicate symbol $``std"$ denoting "standard" to the language
of $\text{ZF}$, now iff we add an omega rule to $``\text {ZF - Infinity + every set is finite}"$ formulated in that language. 
in the form of:
if $\{\varphi_0(y),\varphi_1(y), \varphi_2(y),...\}$ is the set of all formulas in the language of $ZF$ [i.e.; doesn't use the symbol $std$] in one free variable symbol $``y"$, and if $\psi(x)$ is a formula in one free variable symbol $``x"$ that doesn't use the symbol $std$, then
From: for each $n=0,1,2,3,...$; we have:
$\forall x \ [x=\{y|\varphi_n(y)\} \to \psi(x) ]$
We infer:
$\forall x \ [std(x) \to \psi(x)]$
Add the axiom that:
$\forall x \ [x=\{y|\varphi(y)\} \to std(x) ]$, 
for each formula $\varphi(y)$ in one free variable $y$.
Now define "standard natural" as:
$$ \text{standard natural}(n) \iff n  \text{ is a von Neumann ordinal } \wedge std(n)$$
The first question is: Does every standard set have a standard natural as its cardinality? i.e. formally this is:
$\forall x [std(x) \to \exists n (\text{standard natural}(n) \wedge n=|x|)]$
The second question is: can this be generalized over all of $ZFC$, i.e. every standard set in $ZFC$ has a standard von Neumann ordinal as its cardinality.
 A: Regarding your first question: No, this need not be true. Take the standard model $V_\omega$ which must satisfy $\forall x\ \operatorname{std}(x)$. Take a proper elementary endextension $\langle, V_\omega, \in\rangle\prec \langle M, \in\rangle=\mathcal M$. If we let $\operatorname{std}^\mathcal M=M$ then it is easy to see that this structure still satisfies your axiom system (whenever the inference rule tells us that $\forall x\ \operatorname{std}(x)\implies \psi(x)$, we must already have $\forall x\ \psi(x)$ in $V_\omega$ and thus we have $\forall x\ \psi(x)$ in $\mathcal M$ as well). Now take any $x\in M\setminus V_\omega$ that is not an ordinal there and let $n=\operatorname{card}^\mathcal M(x)$. We can define a new structure $\mathcal M^\prime$ by letting $\vert \mathcal M^\prime\vert= M$ but setting $\operatorname{std}^{\mathcal M^\prime}=M\setminus\{n\}$. Since $n$ is not definable in $\mathcal M^\prime$ using a formula without $\operatorname{std}$, $\mathcal M^\prime$ still satisfies this axiom system, however we have $\operatorname{std}(x)$ and $\neg\operatorname{std}(\operatorname{card}(x))$. 
On the other hand, whenever a strucutre $\mathcal N$ satisfies this axiom system, there is a structure $\mathcal N^\prime$ with $\vert \mathcal N^\prime\vert=\vert\mathcal N\vert$, $\in^{\mathcal N^\prime}=\in^\mathcal N$ and $\operatorname{std}^{\mathcal N}\subseteq\operatorname{std}^{\mathcal N^\prime}$ where additionally $\forall x\ \operatorname{std}(x)\rightarrow \operatorname{std}(\operatorname{card}(x))$ holds. Just let $\operatorname{std}^{\mathcal N^\prime}=\operatorname{std}^\mathcal N\cup \{\operatorname{card}(x)\mid \operatorname{std}(x)\}^\mathcal N$.  Since wehenever a set $x$ is definable, its cardinality is definable as well, $\mathcal N^\prime$ will still satisfy this system. 
