Let us take the value of $$a^2 - 6a + 3 = x$$.
Thus, the number of red balls in bag 1 = x+6
The number of red balls in bag 2 = x+4
The number of red balls in bag 3 = x+2
The number of red balls in bag 4 = x
The total balls in Bag 1 = $$3+x+6 = x+9$$
The total balls in Bag 2 = $$5+x+4 = x+9$$
The total balls in Bag 3 = $$7+x+2 = x+9$$
The total balls in Bag 4 = $$9+x = x+9$$
Thus, the probability that the black is selected from each of the bags is as shown:-
Bag 1 =$$dfrac{3}{x+9}$$
Bag 2 =$$dfrac{5}{x+9}$$
Bag 3 =$$dfrac{7}{x+9}$$
Bag 4 =$$dfrac{9}{x+9}$$
The total probability that a black ball will be selected =
$$dfrac{1}{4}$$*($$dfrac{3}{x+9}$$+$$dfrac{5}{x+9}$$+$$dfrac{7}{x+9}$$+$$dfrac{9}{x+9}$$)
= $$dfrac{4}{x+9}$$
The maximum value will occur at $$x=0$$ and that will be $$dfrac{4}{9}$$
Hence, option D is the correct answer.