When $\pu{0.539 g}$ of $\ce{Na(s)}$ reacts with excess $\ce{F2(g)}$ to form $\ce{NaF(s)}$, $\pu{13.3 kJ}$ of heat is evolved at standard-state conditions. What is the standard enthalpy of formation ($\Delta_\mathrm{f}H^\circ$) of $\ce{NaF(s)}$?
I am not asking for the answer. What I would like to know is what "$\pu{13.3 kJ}$ of heat is evolved at standard-state conditions" means. Does anyone know? I am unfamiliar with this terminology.
It simply means that the experiment was carried out under a standard pressure.
This standard pressure can be defined by you to be anything. You may define it to be 1 atm, or 5 atm, or 3 Pa. However, the established convention is to define the standard pressure as $p^\circ = \pu{1 bar}$. See: standard state in the IUPAC Gold Book
What implications does it have for your question? Nothing, really. It just means that the value of $\Delta H$ you derive can be labelled with the word "standard". If you run your reaction at a different pressure $p \neq p^\circ$, then the enthalpy of formation has to be labelled $\Delta_\mathrm{f}H$ instead of $\Delta_\mathrm{f}H^\circ$.
Note that there is no stipulation of temperature. Therefore, the reaction may have been carried out at 10 K, or 1000 K, and it does not have to be 298 K. This simply means that every different temperature will have a different value of $\Delta_\mathrm{f}H^\circ$.