1. Let the event of choosing box $A$ or $B$

Since the box is selected randomly, $P(A)=P(B)=\frac12.$

Let $E$ be the event that in two draws (without replacement), one ball is white and the other is red.

We need to find: $P(B\mid E).$

Using Bayes' theorem, $P(B\mid E)=\frac{P(B)P(E\mid B)}{P(A)P(E\mid A)+P(B)P(E\mid B)}.$

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

Box $A$ contains:

• $2$ white
• $3$ red
• $2$ black

Total balls in box $A$: $7.$

We want probability of getting one white and one red in two draws without replacement.

This can happen in two orders:

• white then red
• red then white

So, $P(E\mid A)=\frac{2}{7}\cdot\frac{3}{6}+\frac{3}{7}\cdot\frac{2}{6}.$

$P(E\mid A)=\frac{6}{42}+\frac{6}{42}=\frac{12}{42}=\frac{2}{7}.$

3. Compute $P(E\mid B)$

Box $B$ contains:

• $4$ white
• $2$ red
• $3$ black

Total balls in box $B$: $9.$

Again, one white and one red can occur in two orders: $P(E\mid B)=\frac{4}{9}\cdot\frac{2}{8}+\frac{2}{9}\cdot\frac{4}{8}.$

$P(E\mid B)=\frac{8}{72}+\frac{8}{72}=\frac{16}{72}=\frac{2}{9}.$

4. Apply Bayes' theorem

$P(B\mid E)=\frac{\frac12\cdot\frac29}{\frac12\cdot\frac27+\frac12\cdot\frac29}.$

Cancel $\frac12$ from numerator and denominator: $P(B\mid E)=\frac{\frac29}{\frac27+\frac29}.$

Now, $\frac27+\frac29=\frac{18+14}{63}=\frac{32}{63}.$

Thus, P(B\mid E)=\frac{2}{9}\cdot\frac{63}{32}=rac{14}{32}=rac{7}{16}.

5. Check options

The correct probability is: $\boxed{\frac{7}{16}}$

So the correct option is B.