Let 9 distinct balls be distributed among 4 boxes $$B_1$$, $$B_2$$, $$B_3$$ and $$B_4$$. If the probability that $$B_3$$ contains exactly 3 balls is $$k\left(\frac{3}{4}\right)^9$$ then $$k$$ lies in the set:

We have 9 distinct balls, and each ball can be dropped in any one of the 4 boxes with equal probability. Thus every ball has 4 independent choices, giving a total of $$4^9$$ equally likely distributions.

We want the event that box $$B_3$$ ends up with exactly 3 balls. To achieve this, we must first select which 3 of the 9 distinct balls enter $$B_3$$. Using the combination formula $$\,^nC_r=\dfrac{n!}{r!(n-r)!}\,$$, the number of ways to make this selection is

$$\binom{9}{3}=84.$$

After fixing those 3 balls in $$B_3$$, the remaining 6 balls are free to go to any of the other three boxes $$B_1, B_2, B_4$$. Each of these 6 balls has 3 choices, so the number of ways to place the remaining balls is $$3^6$$.

Therefore, the total number of favourable distributions is

$$\binom{9}{3}\,3^6.$$

The required probability is hence

$$P=\dfrac{\binom{9}{3}\,3^6}{4^9}.$$

To express this probability in the form $$k\left(\dfrac34\right)^9$$, first note that

$$3^6=\dfrac{3^9}{3^3}.$$

Substituting this inside the expression for $$P$$ we get

$$P=\dfrac{\binom{9}{3}}{3^3}\,\dfrac{3^9}{4^9}= \left(\dfrac{\binom{9}{3}}{3^3}\right)\left(\dfrac34\right)^9.$$

We can now read off

$$k=\dfrac{\binom{9}{3}}{3^3}=\dfrac{84}{27}=\dfrac{28}{9}\approx 3.11.$$

This value of $$k$$ clearly satisfies $$2<k<4,$$ which is exactly the interval described by the set $$\{x\in\mathbb R:\,|x-3|<1\}.$$

Hence, the correct answer is Option A.