NPN transistors are turn ON via current going into the base pin.
First one have to determine the load current. For a lamp of 21W at nominal 12V supply, the current is 21W/12V=1.75A. However, we should allow for extra margins so I would say about 2A.
If you than look at the datasheet graph on page 2 "DC Current Gain", it gives a current gain at this particular current to be 40. Again we allow for some margins so I would say it is about 30. Thus for a collector current of 2A, the base current will be roughly 2A/Gain = 2A/30 = 67mA.
If this is an industrial application, I would even lower the DC gain to 15 according to the figure quoted by page1 of the datasheet(Hfe Min 15 at 3A). This ensure the TIP41C will will be fully ON no matter what.
EDIT: I would look at using a MOSFET instead of a bipolar transistor.
Thus in your case you need to pass a current of 67mA into the base pin to result in a collector current flow of 2A. At this current, the Vbe(sat) is about 1V and you can work out the resistor value say from a 5V drive. Base resistor = (5 - Vbe(sat) ) / 67mA = 60 ohm.
Of course in real life the lamp current will be less than 21W/12V because at 2A, the saturation voltage of the TIP41C is about 2V thus only about 10V is available to light up your lamp. Your TIP41C is dissipating 3~4W of heat so it will definitely need a heatsink.
Hopes this help.