To sum up:
- Increasing C will mainly reduce the cut-off frequency (so at which frequency you start reducing the volume)
- Decreasing R will mainly reduce the volume kept for the highest frequency.
It seems you want to lower some given quite high frequencies, so a .022µF shall be ok (going to .047µF is more about getting a different tone).
So you shall reduce the resistor.
In fact, two resistors in parallel give a lower resistance. Actually, if R1 and R2 are the two initial resistance, you obtain a total resistance that is: R = (R1*R2)/(R1+R2). In particular, if R1=R2, it gives: R = 0.5*R1 .
Note that:
- In order to find the right resistance in you case, you can replace the resistor by a tiny potentiometer (adujstable with a screwdriver). They are cheap, and you can still replace it with a resistor when you know the required value (if you want something proper). Experimenting and using your ear to find your optimum set-up is still the best!
- Just an idea: if you are just bothered by the high-pitched tremolo picking sound, you can even try to increase the cut-off freqency by using a smaller cap. Thus, you could keep more trebbles intact, and still reduce this unbearable picking noise. (Once again: your guitar, your sound. Stick to your ears, not to the "standard".)