Relationship between PSRR and gain
Gain is absolutely the ONLY important part of PSRR. Essentially what you are saying is how much can an op-amp when feeding back a signal cancel out any ripples introduced from the power supply, not from the input of the circuit..

Lets take a simple example: an ideal (infinite open loop gain) voltage follower (output tied directly to the inverting input, fed from the non inverting input). The circuit has a closed loop gain of 1, but the feedback (since the overall gain is SOOO high) will mean that any power supply ripple will be canceled due to the feedback forcing the non inverting and inverting inputs to be in perfect lockstep..

But take the SAME example, but make the OPEN loop gain of the opamp 1, still with closed loop gain of 1, then suddenly the op amp can't keep up with the changes between the non-inverting input and the output-inverting input. And hence all ripple from the power supply would be visible on the output (essentially the op-amp would turn into a noise source with the noise being the coupled power supply ripple)

I understand HOW stevenvh could say that the gain is not meaningful, because he meant CLOSED loop gain... But the gain of question is OPEN loop gain, and YES, that is EVERYTHING in PSRR.

EDIT: And to answer your question, just to follow up slightly here, the PSRR is related to open loop gain, but the more closed loop gain you introduce, the more power supply ripple you will get on the output (hence the 60dB you reference above)

Here is why: Same example I give above, except this time you have a REAL op amp, (finite open loop gain), and resistors in your feedback path, meaning you have a closed loop gain of some value, say 6dB. Since the resistors behave as a voltage divider, the op-amp has to OVERCOMPENSATE for the power supply ripple being fed back to the non-inverting input. If it can only compensate for 100dB of power supply ripple, you will only get 94dB of rejection. The more closed loop gain you introduce, the less of the power supply ripple you are able to reject. 

The whole conversation stems from the separate meanings of open loop and closed loop gain. 

2nd EDIT: And the way that you get 60dB, or I get my 94dB is that you have to realize you have to convert dB BACK so for example you need to use 

$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{6}{20}}\right) = 94 \mathrm{dB}
$$

$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{40}{20}}\right) = 60 \mathrm{dB}.
$$

And YES the other guy who said it should be 1mV not 1µV on Wikipedia is correct.