Suppose there are 4 bags. Bag 1 contains 1 black and $$a^{2}$$ - 6a + 9 red balls, bag 2 contains 3 black and $$a^{2}$$  - 6a + 7 red balls, bag 3 contains 5 black and $$a^{2}$$  - 6a + 5 red balls and bag 4 contains 7 black and $$a^{2}$$  - 6a + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is

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
Bag 4 will have the minimum number of red balls.
The probability of a black ball= $$\frac{1}{4}[\frac{1}{x+7}]+\frac{1}{4}[\frac{3}{x+7}]+\frac{1}{4}[\frac{5}{x+7}]+\frac{1}{4}[\frac{7}{x+7}]$$

In order to maximize the probability we need to minimize the value of x

The minimum value of x can be is 0 as number of red balls cannot be negative. 

Put x=0 we get P(Black ball/total)= 1/4 [16/(x+7)]= 4/7
Hence, option D is the correct answer.