To start, Some of your math is a little off.
For the red LED, if you are using a 5v supply and the red LED needs 20ma and has a voltage drop of 2.1v, then you need a limiting resistorstrong text of (5-2.1)/.02 = 145 ohms.
For the green LED, you need (5-3.2)/.02 = 90 ohms.
For the blue LED, you need (5-3.1)/.02 = 95 ohms.

Assuming that these resistors cause equal LED illumination, and that the light intensity varies directly as the current applied to the LED, then you need to reduce the currents to the green and blue LEDs as follows:

For the green LED, the current needed is (20ma x 20% =) 4ma. For the same voltage drop, the new current limiting resistor required is (5-3.2)/.004 = 450 ohms.
For the blue LED, the current needed is (20ma x 15%) =) 3ma. For the same voltage drop, the new current limiting resistor required is (5-3.1)/.003 = 633 ohms.

Obviously, if the assumptions are not accurate, the calculations will also not be accurate. If more accuracy is required, then you will need to use the LED's data sheet.

To start, some of your math is a little off.
For the red LED, if you are using a 5 V supply and the red LED needs 20 mA and has a voltage drop of 2.1 V, then you need a limiting resistor of (5-2.1)/.02 = 145 Ω.
For the green LED, you need (5-3.2)/.02 = 90 Ω.
For the blue LED, you need (5-3.1)/.02 = 95 Ω.

Assuming that these resistors cause equal LED illumination, and that the light intensity varies directly as the current applied to the LED, then you need to reduce the currents to the green and blue LEDs as follows:

For the green LED, the current needed is (20ma x 20% =) 4 mA. For the same voltage drop, the new current limiting resistor required is (5-3.2)/.004 = 450 Ω.
For the blue LED, the current needed is (20ma x 15%) =) 3 mA. For the same voltage drop, the new current limiting resistor required is (5-3.1)/.003 = 633 Ω.

Obviously, if the assumptions are not accurate, the result of the calculations will also not be accurate.
If more accuracy is required, then you will need to use the LED's data sheet.

To start, Some of your math is off.
For the red LED, if you are using a 5v supply and the red LED needs 20ma and has a voltage drop of 2.1v, then you need a limiting resistor of (5-2.1)/.02 = 145 ohms.
For the green LED, you need (5-3.2)/.02 = 90 ohms.
For the blue LED, you need (5-2.1)/.02 = 145 ohms.

Assuming that these resistors cause equal LED illumination, and that the light intensity varies directly as the current applied to the LED, then you need to reduce the currents to the green and blue LEDs as follows:

For the green LED, the current needed is (.02 x .2 =) 4ma. For the same voltage drop, the internal resistance is (3.2v/.004 =) 800 ohms, and the new current limiting resistor required is (5-3.2)/.004 = 450 ohms.
A similar process is used for the blue LED. Obviously, if the assumptions are not accurate, the calculations will also not be accurate. If more accuracy is required, then you will need to use the LED's data sheet.

To start, Some of your math is a little off.
For the red LED, if you are using a 5v supply and the red LED needs 20ma and has a voltage drop of 2.1v, then you need a limiting resistorstrong text of (5-2.1)/.02 = 145 ohms.
For the green LED, you need (5-3.2)/.02 = 90 ohms.
For the blue LED, you need (5-3.1)/.02 = 95 ohms.

Assuming that these resistors cause equal LED illumination, and that the light intensity varies directly as the current applied to the LED, then you need to reduce the currents to the green and blue LEDs as follows:

For the green LED, the current needed is (20ma x 20% =) 4ma. For the same voltage drop, the new current limiting resistor required is (5-3.2)/.004 = 450 ohms.
For the blue LED, the current needed is (20ma x 15%) =) 3ma. For the same voltage drop, the new current limiting resistor required is (5-3.1)/.003 = 633 ohms.

Obviously, if the assumptions are not accurate, the calculations will also not be accurate. If more accuracy is required, then you will need to use the LED's data sheet.