Deciding On A Resistor For an LED
200 mA will kill it in short order. That's the Absolute MAX.

100 mA and 65 mA are the nominal recommended max currents.

Formula for resistor is:


  (V Source - V LED Forward Voltage) / Target Forward Current
  (Vs - VF) / If


( 5V - 3V ) / 0.1A = 20Ω

22Ω gives slightly less than 100 mA (2V / 22Ω = ~0.09A) will really not be noticeably dimmer, and will help ensure the stated life of the led is met. A target of 65mA might be better. 2V / 0.065A = ~30Ω. As @Asmyldof points out, enclosing the led in a tube, on thin wires, isn't the typical application the manufacturer expects (on a pcb board with adequate landing pads/copper sizing, and certain free air/distance).

Since you have 2 volts across the resistor at 90mA, that's 2 * 0.09 = 0.18 Watts. You will need a 1/4th Watt resistor. So smd resistors are right out unless you go with two 44 ohm 1206 (1/8th Watt) resistors in parallel. So stick to 1/4th or 1/2 Watt through hole resistors.

Keep in mind, the original instructable used a 0805 20mA led. You want a 3535 led, thats normally 5 times bigger light area and 5x more current. It will be particularly bright and blinding in comparison. You will want to play around with one to find out a good brightness. Use a few resistors to test. Between 30Ω and 200Ω before finalizing which one. And the more current used, the faster your usb power pack will drain. At 50mA you will have twice the battery life of 100mA, generally speaking.