You didn't mention how much current the LED needs. Let's assume a standard small LED that requires 20 mA and drops 2 volts....
When the transistor is on (saturated), the emitter will be very near 24 volts. If the LED needs 20 mA and drops 2 volts, then the resistor in the emitter will need to be R = E/I = (24 - 2)/.02 = 1.1K. Use a 1.1K, 1W resistor.
Now, we need to add a resistor to the base drive. We need to provide current to the base to turn the transistor on, but we don't want to cram the entire 52 volts there. If we have 24 volts on the emitter (with the transistor on), then the base will need to be at about 24.6 volts. To fully saturate a BJT, rule of thumb is to provide the base with 1/10 the collector (and emitter) current. So we need 2 mA of base drive.
We have (52 - 24.6) volts of drop across the base resistor, and need it to provide 2 mA, so.. R = E/I = 27.4/.002 = 13.7K. I'd suggest adding a 15K, 1/4W resistor in series with the transistor's base.