For $K=1,2,3$ the box $B_k$ contains $k$ red balls  and $\left(k+1\right)$ white balls. Let $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{3}$ and $P\left(B_3\right)=\frac{1}{6}$. A box is selected at random and a ball is drawn from it. If a red ball is drawn, the probability that it has come from box $B_2$is

The bag $B_1$ contains 1 red balls 2 white balls, so that $P\left(\frac{R}{B_1}\right)=\frac{1}{3}$

The bag $B_2$ contains 2 red balls and 3 white balls, so that $P\left(\frac{R}{B_2}\right)=\frac{2}{5}$

The bag $B_3$ contains 3 red balls and 4 white balls, so that $P\left(\frac{R}{B_3}\right)=\frac{3}{7}$

Hence, by Bayees Theorem 

$\begin{array}{rcl}P\left(\frac{B_2}{R}\right)& =& \frac{P\left({\displaystyle \frac{R}{B_2}}\right)P\left(B_2\right)}{P\left(\frac{R}{B_1}\right)P\left(B_1\right)+P\left(\frac{R}{B_2}\right)P\left(B_2\right)+P\left(\frac{R}{B_3}\right)P\left(B_3\right)}\\ & =& \frac{{\displaystyle \frac{2}{5}}·{\displaystyle \frac{1}{3}}}{\frac{1}{3}·\frac{1}{2}+\frac{2}{5}·\frac{1}{3}+\frac{3}{7}·\frac{1}{6}}\\ & =& \frac{{\displaystyle \frac{2}{15}}}{{\displaystyle \frac{1}{6}}+{\displaystyle \frac{2}{15}}+{\displaystyle \frac{3}{42}}}\\ & =& \frac{2·14}{35+28+15}\\ & =& \frac{28}{78}\\ & =& \boxed{\frac{14}{39}}\end{array}$