1. Define events

Let:

• $E_i$ = event that bag $B_i$ is selected, for $i=1,2,3$
• $W$ = event that the drawn ball is white

Since one of the three bags is selected at random, $P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}$

2. Compute probability of drawing a white ball from each bag

• For $B_1$: 6 white, 4 blue, total 10 $P(W\mid E_1)=\frac{6}{10}=\frac{3}{5}$

• For $B_2$: 4 white, 6 blue, total 10 $P(W\mid E_2)=\frac{4}{10}=\frac{2}{5}$

• For $B_3$: 5 white, 5 blue, total 10 $P(W\mid E_3)=\frac{5}{10}=\frac{1}{2}$

3. Find total probability of getting a white ball

By the law of total probability, $P(W)=P(E_1)P(W\mid E_1)+P(E_2)P(W\mid E_2)+P(E_3)P(W\mid E_3)$

So, $P(W)=\frac{1}{3}\cdot\frac{3}{5}+\frac{1}{3}\cdot\frac{2}{5}+\frac{1}{3}\cdot\frac{1}{2}$

$P(W)=\frac{1}{5}+\frac{2}{15}+\frac{1}{6}$

Taking LCM $30$, $P(W)=\frac{6}{30}+\frac{4}{30}+\frac{5}{30}=\frac{15}{30}=\frac{1}{2}$

4. Apply Bayes' theorem

We need $P(E_2\mid W)=\frac{P(E_2)P(W\mid E_2)}{P(W)}$

Substitute values: $P(E_2\mid W)=\frac{\frac{1}{3}\cdot\frac{2}{5}}{\frac{1}{2}}$

$=\frac{2}{15}\cdot 2=\frac{4}{15}$

5. Match with options

$\boxed{\frac{4}{15}}$

So the correct option is B.