In Hungerford's Algebra, it is written: as anyone familiar with the paradoxes of set theory might suspect, the class of all algebraic extension fields of $K$ need not be a set.
I guess it means that this requires putting all the algebraic extensions of $K$ together, and then construct Russell's paradox.
I am struggling to show the paradox.
On
All groups of order $1$ are isomorphic. But one can take any set $S\in \mathbb V$ in the cumulative hierarchy and turn $S$ into a group of order $1$, producing a vast collection of such groups. Such a collection is not a set for the same reason the cumulative hierarchy is not. Similar reasoning applies to algebraic extensions.
Suppose $K$ has a proper algebraic extension $L$. Let $a\in K\setminus L$. Now for any set $x$, we can form an an extension $L'$ of $K$ isomorphic to $L$ by replacing the element $a$ by $x$ (and defining the field operations so that $x$ adds and multiplies with other elements just how $a$ did).
Now suppose the class $A$ of all algebraic extensions of $K$ is a set. Then $\bigcup A = \bigcup_{F\in A} F$ is a set (by the axiom of union). Since the extensions $L'$ we constructed above are in $A$, the set $\bigcup A$ contains every set, i.e. it is a universal set. But we can prove (e.g. by Russell's Paradox) that there is no universal set.