LED series resistor
The answer is the nearest value you can actually buy, rounded up so the LED never runs over its rating, and the current that value really gives is printed beside it.
300 Ω
(5 − 2) ÷ 10 mA = 300 Ω, up to 300 Ω
Ideal
300 Ω
Really gives
10.0 mA
Resistor burns
30 mW
A whole backlight is this sum in a grid — the LCD backlight driver.
300 Ω gives it 10.0 mA, not the 10 you asked for, because 300 Ω is the nearest part above the ideal 300 Ω and rounding the other way would run the LED over its rating. The resistor burns 30 mW — a 1/8 W part is fine — and the LED itself 20 mW.