30 minutes is half an hour. 323.344 Watts is approximately 324 Watts.
If you want 324 Watts for half an hour, then you want a (324 * 0.5) Watt-hour battery. This assumes, though, that your battery will go from 100% charge to 0% charge after exactly 30 minutes.
Discharging a battery down to a state of charge of 0% can physically damage batteries, preventing them from ever charging again, and the battery voltage will usually fall quite dramatically as the state of charge approaches 0%, so you might wind up browning out your electronics before you actually hit 0%, leaving the last few percent of state of charge uncaptured and thus falling short of your desired 30 minutes of run time.
What might be a better approach would be to assume that you want your batteries sized such that you can supply the full (324*0.5) Watt-hours of capacity when the batteries are discharged from, for example, 80% of full charge down to 20% of full charge. That is, you want:
$$
(0.80 - 0.20) * \mbox{Capacity} = 324*0.5 \mbox{Watt-hours}\\
$$
The difference between your starting and ending state of charge is called, not surprisingly, the change or delta ($\Delta$) state of charge, and you can solve for capacity by:
$$
\Delta \mbox{SOC} * \mbox{Capacity} = (324 * 0.5) \mbox{Watt-hours}\\
$$
$$
\mbox{Capacity} = \frac{324*0.5}{\Delta\mbox{SOC}} \mbox{Watt-hours} \\
$$
In the example I gave, where you are designing your system to supply your desired capacity between 80 and 20 percent of full state of charge, your $\Delta\mbox{SOC} = 0.8-0.2 \boxed{= 0.6} $, so you wind up with a target battery capacity of:
$$
\mbox{Capacity} = \frac{324*0.5}{0.6} \mbox{Watt-hours} \\
\boxed{\mbox{Capacity} = 270 \mbox{Watt-hours}} \\
$$
Remember that $P = IV$, so 270 Watt-hours could be rewritten 270 volt-amp-hours. You can divide your target capacity by the nominal battery voltage to get the actual battery size in amp-hours.
