Do I use a 5V 2A or 3A DC adapter to power my 2m (6.5ft) WS2801 LED strip?
Current ability of supply should be equal or greater than current required.
A supply able to supply more than required will do no harm.
As I = V/R , current is provided "on demand". The supply supplies what is needed, not more.

For reasons explained below, a 3A supply is just about right for your 2m strip.
A 2A supply is underrated and a >3A supply will do no harm.

In the calculations below I have used the figures to the number of places specified and not rounded the results in calculations. In practice the values in the specification sheet will be approximate and will vary between samples. A variation of +/- 5% in currents would be good, +/- 10% not surprising and > 10% not unknown.
Such variations will usually not matter. Occasionally they will.
It pays tp provide somewhat more power supply capacity than required as the voltage may "sag" somewhat when loaded to the limit. Some supplies will get excessively hot when at or slightly above their specified ratings BUT these are liable to also have other problems - such as being death traps occasionally. A good 3A supply from a competent manufacturer will usually give you 10% + above the rated value without demur. 



Your specifications are inconsistent
Current = Power/Volts.  or I = P/V
If the strip was 7.68 W/m as claimed then at 1m the current would be
I = P/V = 7.68/5 = 1.556A and 5m = 7.78A

Suspect values:

If I = 3.2A (another of your specs) then with 32 LED/s each LED would draw 3.2A/32 = 100 mA. This is conceivable with all 3 LEDs operating - some suppliers say 60 mA = 20 mA per LED colour and 100 mA = 33 mA per LED colour. It is usual but not certain to rate LEDs of this size at 20 mA / LED colour - but it is possible to be higher.

At 3.2A/metre 1m will take 5V x 3.2A = 16W
and 5m will take 5 x 16W = 80W
which seems 'rather high' [tm].

Probable correct values:

If we instead use the quoted 7.68W/m
I/m = P/V = 7.68/5 = 1.536A
As there are 32 LEDS/m this means
each LED takes 1.5436/32  = 48 mA/LED
or 16 mA per LED segment.
This sounds closer to likely reality (but may not be).

Assume the 3.2A figures is correct. Then:

V = 5V
I per m = 1.536A
I per 5m = 7.68A &lt;- explains the 7.68 figure in the data.
P per m = 7.68W
P per 5m = 38.4 W

2m requires 2 x 1.536 = 3.072A

ie a 3A supply is about right.



M for metre should be m as Mr Metre did not exist.