I agree with your answer Radarcontact, but your approach to solving the problem only works in simple cases and may not be the "approved" method. The way I'd treat it is:-
Taking moments about the mainwheel, there are two to consider. 20,000N at 0.5m and ftw N at 8m. Since the aircraft is not rotating, these sum to zero. Thus:-
0 = (-20,000 x 0.5) + (8ftw)
or
ftw = (20,000 x 0.5) / 8 = 1250 N, which is the force at the tailwheel.
Similarly, taking moments about the tailwheel, there are again two to consider. 20,000N at 7.5m, and fmw at 8m. Since the aircraft is still not rotating, these still sum to zero, hence:-
0 = (20,000 x 7.5) + (-8 x fmw)
or
fmw = 20,000 x 7.5 / 8 = 18,750
Assuming that the load on the mains is evenly distributed between them, divide by 2 to get 9,375 N per mainwheel.
And a crude check that you got it right is to add all three up and confirm that they add up to 20,000, which is the weight of the aeroplane.
G