Graham's Law of Effusion:

The rate of effusion of a gas is inversely proportional to the molar mass of the gas.

rate of gas 1 / rate of gas 2 = sqrt MW gas 2 / sqrt MW gas 1

Since we are only being asked to COMPARE the rates of CO_{2} and O_{3}, we can just choose an arbitrary rate for O_{3} as 1 and compare the rate of CO_{2} to that.

rate of O_{3} = 1

MW O_{3} = 48

rate of CO_{2} = x

MW CO_{2} = 44

1/x = sqrt 44 / sqrt 48

1/x = 0.917

x = 1.09

So, this tells us that CO_{2} will diffuse 1.09 times faster than O_{3}, or it will diffuse 9% faster