The data sheet states that with the LED dissipating 2.63 watts, its forward voltage, Vf, will be 3.5 V, which corresponds to a forward current, If, of 751 mA.
With the LED dissipating 3.38 watts, its Vf will be 4.5V and its If also 751 mA.
That says that with 751 mA through the LED its Vf can lie anywhere between 3.5 and 4.5 volts, a caveat being that the LED's absolute maximum If is 750 mA.
Since the LEDs are going to be used as aquarium lights, there's probably no reason to blind the fish and run the LEDs on the hairy edge of their own demise.
The If VS Vf graph on Ebay, for the LED, shows that with an If of about 400 mA, Vf will be about 3.5 V, allowing the LEDs to dissipate a much more comfortable 1.4 watts each.
Going that way means that you can wire three LEDs in series for a total of 10.5 volts across the string, and drop the remaining 1.5 volts from the 12 volt supply with a 3.75 ohm resistor in series with the string. The resistor will dissipate 0.6 watts, so a 1 or two watt resistor would be a good choice.
A nice metal oxide 3.6 ohm 2 watt unit is readily available, and would up the current through the LEDs a little, but would be fine for the application.
So, since you can't get 4 LEDs in series to be properly driven by 12 volts, the array changes from 4 X 5 to 3 X 6 with two LEDs left over.
If the 3 X 6 array would work for you, then the supply would have to put out 2.4 amps at 12 volts, which is about 29 watts, well within its capabilities.
If you absolutely have to use all 20 LEDs, then you could connect the last two in series with a 12.5 ohm 3 watt resistor and wire that string in parallel with all the others, increasing the load on your supply to about 34 watts.