Let

• $E$ = event that the chosen box is $B_2$
• $A$ = event that in drawing $2$ balls, one is white and one is red

We need to find: $P(B_2\mid A)$

Since the box is selected randomly, $P(B_1)=P(B_2)=P(B_3)=\frac13$

We use Bayes' theorem: $P(B_2\mid A)=\frac{P(B_2)P(A\mid B_2)}{P(B_1)P(A\mid B_1)+P(B_2)P(A\mid B_2)+P(B_3)P(A\mid B_3)}$

Because all three prior probabilities are equal, this becomes $P(B_2\mid A)=\frac{P(A\mid B_2)}{P(A\mid B_1)+P(A\mid B_2)+P(A\mid B_3)}$

1. Compute $P(A\mid B_1)$

Box $B_1$ has $1$ white, $3$ red, $2$ black balls, total $6$.

Probability of getting one white and one red: $P(A\mid B_1)=\frac{\binom11\binom31}{\binom62}=\frac{1\cdot 3}{15}=\frac15$

2. Compute $P(A\mid B_2)$

Box $B_2$ has $2$ white, $3$ red, $4$ black balls, total $9$.

$P(A\mid B_2)=\frac{\binom21\binom31}{\binom92}=\frac{2\cdot 3}{36}=\frac16$

3. Compute $P(A\mid B_3)$

Box $B_3$ has $3$ white, $4$ red, $5$ black balls, total $12$.

$P(A\mid B_3)=\frac{\binom31\binom41}{\binom{12}2}=\frac{3\cdot 4}{66}=\frac{2}{11}$

4. Apply Bayes' theorem

So, $P(B_2\mid A)=\frac{\frac16}{\frac15+\frac16+\frac{2}{11}}$

Now add the denominator: $\frac15+\frac16+\frac{2}{11}$ LCM of $5,6,11$ is $330$.

$\frac15=\frac{66}{330},\quad \frac16=\frac{55}{330},\quad \frac{2}{11}=\frac{60}{330}$

Hence, $\frac15+\frac16+\frac{2}{11}=\frac{66+55+60}{330}=\frac{181}{330}$

Therefore, $P(B_2\mid A)=\frac{\frac16}{\frac{181}{330}}=\frac16\cdot\frac{330}{181}=\frac{55}{181}$

5. Final answer

$\boxed{\frac{55}{181}}$ This corresponds to Option D.