Finding the Amp hour of battery

The setup in your answer is a bad one.

First you parallel two batteries, which you shouldn't, because their voltages are never exactly the same. Their low internal resistance will cause a current from one battery to the other. So feed three LEDs from 1 battery, and the other three from the other.

Then you place all LEDs parallel, which means that you lose (12 V - 3.2 V) x 20 mA = 176 mW per LED in its series resistor, while the LED itself uses only 64 mW. Total power loss is more than 1 W. That's because the large voltage difference between battery and LED. The best way to get a longer endurance is to keep losses in the series resistors as low as possible. So better place two times three LEDs in series, so that their total current is 40 mA instead of 120 mA. The power loss in the series resistors is then (12 V - 3 x 3.2 V) x 20 mA = 45 mW per 3 LEDs, or 96 mW in total. That's less than 10 % of the power loss for all LEDs in parallel.

Then your batteries will last 100 mAh / 40 mA = 2.5 hours or 150 minutes. This is pretty optimal. The batteries' capacity is 1200 mWh, and the LEDs consume 384 mW, so with an ideal conversion you can get a little over 3 hours out of them. But the most efficient conversion using a switching current regulator will get you maybe 85 % efficiency, and then you only gain 9 extra minutes.

edit re comments
An alkaline battery's voltage quickly drops by 10-15 %, and then remains more constant for a great part of the discharge cycle. So either you calculate the resistors for a larger current at the start, and 20 mA for the rest, or for 20 mA at the start, and a lower current later on. The latter solution will give you a longer battery life, but a bit less brightness.

jippie suggests to use a switcher anyway to get more out of the batteries, and it's a thought. You'll have to place the batteries in series to get 24 V to allow a voltage drop as high as possible. The larger Vin/Vout ratio of the switcher will make it less efficient, but overall you should get some extra time from the batteries.

Your hunch that batteries have a current limitation is correct. In general, it's hard to tell the current rating [A] from capacity [A·h]. You have to look it up in the datasheet. A lot depends on the design of the battery.

For example: coin cells with 500mAh capacity may have only 3mA max current.
Another (opposite) example: automotive starter battery with 40Ah capacity may have 500A max current.

Lead-acid batteries are interesting in this respect, because there are two distinct types.

1. Starter lead-acid batteries are designed specifically to deliver high peak current for a short period of time. Deep discharge, however, dramatically shortens the life of a starter battery. So, it's not suited for routine operation at high depths of discharge. Your typical starter battery in the automobile works at very shallow depth of discharge.
2. Deep cycle lead-acid batteries are designed (as name suggests) to discharge further. But they can not provide as much instantaneous current.
   Here's an example datasheet for a deep cycle battery. Have a look at the nominal capacity on p.1. Notice that capacity depends on discharge current (i.e. the rate of discharge).

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Depth of Discharge    Starter Battery    Deep-cycle Battery

100%                  12–15 cycles       150–200 cycles

50%                   100–120 cycles     400–500 cycles

30%                   130–150 cycles     1,000 and more cycles

(Source.)

p.s. If you want to read-up, here's an excellent web site on batteries - Battery University.