SUPPOSE $$M_{(base-n)}$$ Means m is a number in base-n system. It for $$a > b, a_{(base-10)} + b_{(base-10)} = 20_{(base-3)}$$ and $$a^{2}_{(base-10)} + b^{2}_{(base-10)} = 202_{(base-3)}$$, then what are $$a_{(base-3)}$$ and $$b_{(base-3)}$$?

We need to first convert $$20_{\left(base-3\right)}$$ and $$202_{\left(base-3\right)}$$ into decimal system so as to get the values of a and b in the decimal system.
20 in 10 base system would be : $$\left(2\times\ 3^1\right)+\left(0\times\ 3^0\right)$$ = 6
202 in 10 base system would be : $$\left(2\times\ 3^2\right)+\left(0\times\ 3^1\right)+\left(2\times\ 3^0\right)=20$$

so a+b = 6 and $$a^2+b^2=20$$
From this we get, ab = 8
Trying out different values we can get the values of a and b to be 4 and 2 
converting these from decimal to 3 base system we get 2 as 2 only and 4 as $$\left(1\times\ 3^1\right)+\left(1\times\ 3^0\right)$$ = $$11_{\left(base-3\right)}$$

Therefore, Option A is the correct answer.