(Voltage of LEDs x How many LEDs = Voltage Drop) White LEDS 3.6x3=10.8 then (Power Supply Voltage - Voltage Drop = additional Voltage drop) 12v-10.8=1.2
(additional Voltage drop / milliamp = ohms) 1.2/.025ma =48 ohms
As an example, for a 12 volt light, you can run a maximum of 3 white LEDs in series at full power (3.6 x 3 = 10.8 volts drop). Subtract this from your supply voltage of 12 volts to get the additional voltage that must be dropped (in this case, 12 - 10.8 = 1.2 volts of additional drop needed). In this case, 1.2 volts of additional drop / .025 amps (25 ma) = 48 ohms. Use the next highest value of resistor available, 50 ohms. You must also be sure the resistor can handle enough current. Volts x Amps = Watts; resistors are rated in watts. So in this case, 1.2 volts x .025 amps = 0.03 watts. A 1/4 watt resistor will work fine, but if you run a second string of 3 LEDs in parallel, each string would need its own 50 ohm resistor. It's important that each string has its own resistor....putting them in parallel with a single resistor is bad practice.
http://www.otherpower.com/otherpower_lighting_leds.html
(https://s16.postimg.cc/xm0fx4dfl/LED_Strip_Sizes.png) (https://postimg.cc/image/xm0fx4dfl/)