At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 for complete combustion.

The volume of O2 in 375 mL air containing 20% O2 is (20/100) * 375 mL = 75 mL.

Let the hydrocarbon be CxHy. The balanced combustion reaction is:
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O

From the reaction, the volume ratio of hydrocarbon to O2 is 1 : (x + y/4).

Given that 15 mL of hydrocarbon requires 75 mL of O2, we have:
15 mL / 75 mL = 1 / (x + y/4)
1/5 = 1 / (x + y/4)
x + y/4 = 5
4x + y = 20

Now, let's check the options:

For C3H8:
4(3) + 8 = 20 (This satisfies the equation)

For C4H8:
4(4) + 8 = 24

For C4H10:
4(4) + 10 = 26

For C3H6:
4(3) + 6 = 18

Only C3H8 satisfies the equation 4x + y = 20. Therefore, the gaseous hydrocarbon is C3H8.