Optics question

Hi guys

I'm flicking thorugh some past papers for an upcoming exam and came across this seemingly easy problem.

"The average intensity of solar radiation at the Earth is 1.4kW/m^2. Assuming the earth is a perfect absorber, calculate the force exerted by the radiation on the surface of the Earth. The mean radius of the earth is 6400 km"

Now I know that the radiation pressure on a perfect absorber is equal to the energy density, which is equal to the intensity divided by the speed of light.

$$P = U = \frac{I}{c}$$

Given that P=F/A I get that the force should be.

$$F = \frac {IA}{c} = \frac {I 4\pi R^2}{c}$$

Although it could be $$\frac {I 2\pi R^2}{c}$$

As the sunlight only directly hits approximately half of the Earth's surface. But either way, when I plug in the numbers I don't get the answer my Lecturer gave of 600 MN.

I must be missing something, but I can't see it. If any of you guys can I'd appreciate it.