Given,

In bag $1\to 3R, 4B \& 3W$

And in bag $2\to 2R, 5B \& 2W$

Let $A:$ Drawn ball from bag $\mathrm{II}$ is black

And let $B:$ Red ball transferred

Now by total probability theorem we have $P\left(A\right)=\frac{3}{9}\times \frac{5}{10}+\frac{4}{9}\times \frac{6}{10}+\frac{3}{9}\times \frac{5}{10}=\frac{54}{90}$

And $P\left(A\cap B\right)=\frac{3}{9}\times \frac{5}{10}$

So, by bayes theorem we have, $P\left(\frac{B}{A}\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

$=\frac{\frac{3}{9}\times \frac{5}{10}}{\frac{3}{9}\times \frac{5}{10}+\frac{4}{9}\times \frac{6}{10}+\frac{3}{9}\times \frac{5}{10}}$

$=\frac{15}{15+24+15}=\frac{15}{54}=\frac{5}{18}$