The number of ways in which $9$ distinct balls can be distributed among $4$ boxes is $4^9$

The number of ways in which $B_3$ contains $3$ balls is ${}^9\mathrm{C}_3·3^6$

Using the classical definition of probability we can say the required probability $=\frac{{}^9\mathrm{C}_3·3^6}{4^9}$

$=\frac{{}^9\mathrm{C}_3}{27}·{\left(\frac{3}{4}\right)}^9$

$=\frac{28}{9}·{\left(\frac{3}{4}\right)}^9\Rightarrow k=\frac{28}{9}\approx 3.11$

Now, consider $|x-3|<1$

$\Rightarrow -1<x-3<1$

$\Rightarrow -1+3<x<1+3$

$\Rightarrow 2<x<4$

$\therefore k$ satisfies $|x-3|<1$