I am sure for a trial it does not need to be a big heat sink finned one like that..OK for constant use, yes, to save it burning out, but to test out the theory, you need to know what the actual value the flasher you have is designed for,
As the e-bay advert says,,that value resistor is OK for a system designed for 12volt 21 Watt bulbs.
Assuming two bulbs in the original system, that would be 42 watt
P=IV
42 = I x 12
so current = 42/12 = 3.5 amp current through the the relay flasher, and this is what the resistor must draw to make that flasher unit work
V=IR
or in this case R=V/I = 12 /3.5 =3.4 so I reckon that resistor they are selling is a 3.4 Ohm resistor rated at 50 Watt
But that resistor is only suitable for a flasher that was designed for those particular wattage bulbs.
Need to know what value YOUR actual flasher unit was designed for.
if you do not know, then I suggest finding a variable resistor (potentiometer) of probably ...umm I don't know...say 47 Ohm..something liek that.
put that across the LED, and with it started at max resistance, slowly reduce resistance till the flasher unit starts flashing...
if it is a low power pot, then it probably will get hot quickly, as I could be drawing a lot of current, 2 , 3 4 amps...till it burns up..So so not leave it running. Just see that the flasher works and flashes the LED.
Stop then measure the resistance of the pot and ONLY then buy a resistor of a suitable value.
Of if you have old scrap PSU's kicking around,scavenge old resistors off them,and parallel up as many as needed to get resistance down so the flasher works