= Solution
Let $\mathcal O$ be the set of strict total orders on $\kappa$. Each is a subset of $\kappa\times\kappa$, so the <square of an infinite cardinal> gives
$$
|\mathcal O|\leq2^{|\kappa\times\kappa|}=2^\kappa.
$$
For the lower bound, use the set $\kappa\times\{0,1\}$ and order the two-element blocks by their first coordinates, using the <well-order> of $\kappa$. For every $A\subseteq\kappa$, put $(\alpha,0)$ before $(\alpha,1)$ exactly when $\alpha\in A$, reversing them otherwise. This defines a strict total order for each $A$, and different subsets give different orders. Transport these orders to $\kappa$ along a fixed <bijection> $\kappa\times\{0,1\}\cong\kappa$. Thus $2^\kappa\leq|\mathcal O|$, and the <Cantor-Schröder-Bernstein theorem> gives
$$
\boxed{|\mathcal O|=2^\kappa.}
$$
If using the customary <axiom of choice> development of <infinite cardinal arithmetic>, its use here is precisely to supply $\kappa\times\kappa\cong\kappa$ and $\kappa\times2\cong\kappa$. The ordering construction makes no further choices. It proves the <cardinality of the family of strict total orders>. For $\kappa$ itself an <initial ordinal>, these particular product identities can also be proved in <ZF> by well-ordering pairs according to their maximum coordinate, so full <axiom of choice> is stronger than this instance intrinsically requires.
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