1. Understand the transfer cases

From Bag 1:

• White balls = $4$
• Black balls = $5$
• Total = $9$

So, the probability that the transferred ball is:

• White = $\frac{4}{9}$
• Black = $\frac{5}{9}$

Bag 2 initially has:

• White balls = $n$
• Black balls = $3$
• Total = $n+3$

After transfer, Bag 2 will have total $n+4$ balls.

2. Case-wise probability that the final drawn ball from Bag 2 is white

Case 1: A white ball is transferred

This happens with probability $\frac{4}{9}$.

Then Bag 2 has:

• White = $n+1$
• Black = $3$
• Total = $n+4$

So, $P(\text{white from Bag 2} \mid \text{white transferred}) = \frac{n+1}{n+4}$

Case 2: A black ball is transferred

This happens with probability $\frac{5}{9}$.

Then Bag 2 has:

• White = $n$
• Black = $4$
• Total = $n+4$

So, $P(\text{white from Bag 2} \mid \text{black transferred}) = \frac{n}{n+4}$

3. Use total probability

Given probability of drawing a white ball from Bag 2 is $\frac{29}{45}$:

$\frac{4}{9}\cdot \frac{n+1}{n+4} + \frac{5}{9}\cdot \frac{n}{n+4} = \frac{29}{45}$

Take common denominator:

$\frac{1}{9(n+4)}\left[4(n+1)+5n\right] = \frac{29}{45}$

$\frac{1}{9(n+4)}(4n+4+5n) = \frac{29}{45}$

$\frac{9n+4}{9(n+4)} = \frac{29}{45}$

Cross-multiply:

$45(9n+4)=29\cdot 9(n+4)$

$405n+180=261n+1044$

$405n-261n=1044-180$

$144n=864$

$n=6$

4. Check options

• A: $5$ ✗
• B: $6$ ✓
• C: $4$ ✗
• D: $3$ ✗

So the correct option is B.