You may also be able to put 2 LEDs in series with a single resistor with a 5 V source (depends on the LED voltage drop). That would cut your current draw in half.

Making some assumptions:

17mA current flow desired

Vce - 0.5V,voltage drop across the transistor when on

Vled = 2V, voltage drop across the LED when on

Vsource = 5V, as shown on your schematic

Applying Ohms law, V=IR or V/I = R to calculate Rlimit:

(Vsource - Vled - Vled - Vce)/current = Rlimit

(5V -2V - 2V - 0.5)/.017 = 29.4ohm. so use a 30 or 33 ohm resistor

So, look into your specs for the LED and transistor to be used and calculate the current limit resistors needed.

Power rating needed will be small - with 0.5V across the resistor & 17mA, the resistor is only dissipating 8.5mA (P=IV, so 0.017 * 0.5), so 1/8W resistors are fine.