I have never used an arduino in my life, but I need help. I am trying to create a system of blacklights that can be plugged in and controlled to brighten/dim/turn off/on on command, using an arduino. I have concluded that the best arduino to use for this system is an arduino due. However, I do not know what kind of resistor will work in this system.

I know the resistor needs to have 220 ohm. Could someone tell me if it needs to be made out of carbon, how many watts it needs to have, etc? I would really appreciate it.

There are four variables in linear electronics like these you need to know:

V = Volts (V)
I = Current (A)
R = Resistance (?)
P = Power (W)

The formulae that relate them all are:

You might want to print that wheel out and pin it on your wall.

So the resistance you must have calculated with R=V/I, where V is the output voltage minus the LED forward voltage, and the current is the LED forward current. Say a 20mA and 2.4V LED. That's 3.3-2.4 = 0.9V at 20mA which is 45?. Not sure where you got 220? from.

The power rating, we can use any of the power formulae for as we have all three values now. Simplest is V×I, so 0.9 × 0.02 is 0.018W.

As for what the resistor is made of... a small wattage (1/4W) resistor that's not carbon would be hard to find. You don't need to worry about that kind of thing unless you're working with larger powers or ultra high precision. Just get the normal 1p each resistors.

Then you know wrong, that is just a ball park figure, the real value depends on the actual voltage and LEDs used, but 220R is good enough in most cases.

.Could someone tell me if it needs to be made out of carbon, how many watts it needs to have, etc? I would really appreciate it

It matters not what it is made of. Carbon is cheaper and that is good enough for an LED but anything will do.
As to the power, again the LED will cause so little power to be dissipated that you will find it almost impossible to buy a resistor with too little power rating. So 1/4 watt, 1/8 watt or 1/10 watt are all going to work no problem.