The efficiency is an extremely complex topic, and in very simple terms Wil's answer is correct, but...
The test specifies following parameters:
87% efficiency at 20% of stated load,
90% efficiency at 50% of stated load,
87% efficiency at 100% of stated load
and power factor of 0.9 at 50% of load. Voltage level is expected to be 115 V.
How the power factor part goes into this isn't immediately clear.
So what does all of that mean?
Well, let's take a look at test protocol.
From it we learn that the efficiency is considered ratio between real power provided at device output and real power provided at device input. This is important when calculating whole power consumption of the device. From it we can see that the power factor isn't taken into concern when the efficiency is calculated. So at 50% load, we can more or less accurately calculate total power consumption of the device: First, we say it provides 500 W. Then we take efficiency in, so we know that it consumes around 555.56 W of power. At last, we take power factor into equation and we get that it consumes around 617.28 VA. That is the total power consumed by the device. The real power power supply consumes is 555.56 W and it also consumes 61.72 VAr for which you may or may not be charged, depending on the way electricity is sold in your area.
We can't be sure what the total power consumption will be at 100% load, be cause we can't take into effect the power factor which is unknown at the 100% load. I don't think that it will be considerably lower than the on at 50% and it may be higher.
Also voltage level can make an impact at efficiency of the power supply. Power supplies running at 230 V are usually a bit more efficient than when running at 115 V, but the results will vary from PSU to PSU.
Also about the card power consumption. Here, card is considered a separate system and its own efficiency isn't taken into account. Instead, you're provided with its maximum power consumption. So a 200 W card will pull up to 200 W from the power supply. However, if the power supply is 90% efficient at the moment, you could say that the card will pull 220 W form the power plug of the computer plus an unknown number of volt-ampere reactive.