$TC(A)$ is the transitive closure of $A$, $|x|$ is the cardinality of $x$, $\kappa$ is a cardinal number greater than $\aleph_0$ and $H_\kappa$ is the set of sets hereditarily of cardinality less than $\kappa$.
$\{A: |TC(A)|<\kappa \} \subseteq \{A: (\forall x \in TC(\{A\})) (|x|<\kappa)\}$ is obvious.
I could prove the other direction when $\kappa$ is a regular cardinal because for every $A \in \{A: (\forall x \in TC(\{A\})) (|x|<\kappa)\}$, the cardinality of each of $A, \bigcup A, \bigcup\bigcup A, ... $ is $< \kappa$ and the cardinality of their union, which is $TC(A)$, is $< \kappa$ as well.
I couldn't prove this part for singular cardinals.
Good! Cause it's false when $\kappa$ is singular.
Let $A = \{\aleph_0,\aleph_1, \aleph_2, \dots\}$ (i.e. the set of all infinite cardinals $<\aleph_\omega$). Then $TC(\{A\}) = \{A\}\cup A\cup \aleph_\omega,$ so all $x\in TC(\{A\})$ have $|x|<\aleph_\omega.$ But $|TC(A)| = \aleph_\omega.$