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.
According to APC, the original RBC17 battery has "108 Volt-Amp-Hour" (i.e., 9 AH) capacity.
It may also have better deep-discharge characteristics than a generic 7AH battery.
Best Answer
W is a measure of power, therefore it's likely that it's rated in Wh (Watt-hour, energy).
Since Ah represent a charge (current * time) you can multiply it by the voltage to get Wh; conversely, divide the Wh by the voltage to get the Ah rating.
Looking at the datasheet, there seem to be a reason why a single Ah value is not specified: it varies depending on the load current (see table in the bottom).
The 21 W indicates that it can provide 21 W for 15 minutes (21/4=5.25 Wh) when the cell voltage is at least 1.67 V, times 6 cells means 10 V. This voltage is the minimum tolerated value, and influences the depth of the discharge and, in the long term, the lifetime of the battery.
If you look at the table, the battery is approximately 4 Ah.