By axiom of union for any set A there is a set B such that x belongs to B if and only if x belongs to some z which belongs to A.
According to this everything is a set.My question is what would union of {1,2,3} be?
If I am correct it would imply that there is an x which belongs to 1 or 2 or 3?
How is it possible that somwthing belongs to a natural number?
In ZFC set theory, everything is a set, and in order to speak about numbers you need to chose some set to represent each number.
The most commonly used representation of the natural numbers is to choose to represent the number $n$ by the set $\{0,1,\ldots,n-1\}$.
Under this representaion (the Von Neumann ordinals), the number $0$ is represented by the empty set, $1=\{0\}=\{\varnothing\}$, $2=\{0,1\}=\{\varnothing,\{\varnothing\}\}$, and $3=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}$. Note that the set representing each number has exactly as many elements as the number it represents.
Then the union of $\{0\}$, $\{0,1\}$ and $\{0,1,2\}$ happens to be $\{0,1,2\}$ or $3$ itself.
Such an union doesn't immediately appear to be arithmetically meaningful, but it's actually a quite useful operation when working with ordinals -- it produces the least upper bound of a set of ordinals.
In $\sf ZFC$, as said, every object in the universe is a set. Including the objects we use to represent the natural numbers with.
Note, by the way, that this is not a very peculiar idea. We use Dedekind cuts to represent real numbers. In that case $\{1_\Bbb R,2_\Bbb R,3_\Bbb R\}$ is a set of subsets of the rational numbers, and their union is actually $3_\Bbb R$.
If we consider the real numbers as equivalence classes of Cauchy sequences instead then the union is not a real number itself, but rather just a set of sequences of rational numbers with certain properties.
I'm pointing this out, because while we often think about the real numbers as "absolute" and the rest of the number systems defined from them, we also learn in the first couple of weeks of our mathematical degrees that the real numbers can be seen as subsets of the rational numbers instead.
But, back to $\sf ZFC$. In $\sf ZFC$ everything is a set. In particular, $1,2,3$ are all sets. So the question what is $\bigcup\{1,2,3\}$ makes a lot of sense in $\sf ZFC$. The answer depends on many things.
How did you decide to represent $1,2,3$?
Are these real numbers, natural numbers, rational numbers, complex numbers, etc. etc., since in set theory we often begin by defining the natural numbers and construct the rest of these systems from the ground up, it follows that $1_\Bbb N\neq 1_\Bbb Z\neq 1_\Bbb Q\neq 1_\Bbb R$. But that's not new, in many algebraic contexts one thinks about these systems as different with canonical embeddings from one to the other.
The same is true in set theory. Here though, after all the construction you can decide that now you have the real numbers and you forget everything that you had before and define $\Bbb{N,Z,Q}$ as subsets of the real numbers. In that case, the above $\neq$ become $=$.
Perhaps you mean ordinals? Ordinals are canonical objects in set theory, and they have rather concrete representations. In that case, $\bigcup\{1,2,3\}=3$. But these are still sets.
Maybe we go beyond $\sf ZFC$. Maybe we assume the existence of atoms, or urelements, which are not sets. These are some "non sets" objects, and we can assume that they exist. We can even assume that there are $2^{\aleph_0}$ of them and we can fix a structure of the real numbers on them. (Or we can assume there are $\aleph_0$ of them and give them a different structure instead).
In that case, $\{1,2,3\}$ is a set of objects which are not sets. So $\bigcup\{1,2,3\}=\varnothing$. Because none of $1,2,3$ have elements of its own.
But commonly, we consider the natural numbers to be represented by the finite ordinals, and the ordinals to be represented by the von Neumann ordinals. So these are fairly canonical sets, and so $\bigcup\{1,2,3\}=3$.
This is the same representation as in Henning's answer, which explains nicely why the above equality holds.
See Axiom of union :
Given any set $A$, there is a set $B$ such that, for any element $c$, $c$ is a member of $B$ if and only if there is a set $D$ such that $c$ is a member of $D$ and $D$ is a member of $A$.
In symbols :
$\forall A \exists B \forall c(c \in B \leftrightarrow \exists D(c \in D \land D \in A)$.
In your example : $A= \{ 1,2,3 \}$. Which are the elements of $B = \cup A$ ? They are the elements of the elements of $A$, i.e.the elements of $1$ "plus" the elements of $2$ "plus" the elements of $3$.
Of course, if there are no set $D \in A$, $B$ will be the emptyset ($\emptyset$).
See also the discussion in this post.
