1. Total balls in the urn

There are:

• $6$ white balls
• $9$ black balls

So total balls $= 15$.

Two successive draws of $4$ balls each are made without replacement.

We need:

• first draw: all $4$ are white
• second draw: all $4$ are black

2. Probability that first draw gives 4 white balls

Out of $15$ balls, we draw $4$ balls.

Number of ways to draw any $4$ balls: $\binom{15}{4}$

Number of favorable ways to draw $4$ white balls from $6$ white balls: $\binom{6}{4}$

So, $P(\text{first draw all white}) = \frac{\binom{6}{4}}{\binom{15}{4}}$

3. Probability that second draw gives 4 black balls given first draw was all white

After drawing $4$ white balls, remaining balls are:

• white: $6-4=2$
• black: $9$

So remaining total balls $=11$.

Now second draw is of $4$ balls from these $11$ balls.

Number of ways to draw any $4$ balls: $\binom{11}{4}$

Number of favorable ways to draw $4$ black balls from $9$ black balls: $\binom{9}{4}$

Thus, $P(\text{second draw all black} \mid \text{first draw all white}) = \frac{\binom{9}{4}}{\binom{11}{4}}$

4. Multiply the probabilities

Required probability: $P = \frac{\binom{6}{4}}{\binom{15}{4}} \cdot \frac{\binom{9}{4}}{\binom{11}{4}}$

Now compute values: $\binom{6}{4}=15, \quad \binom{15}{4}=1365$ $\binom{9}{4}=126, \quad \binom{11}{4}=330$

So, $P = \frac{15}{1365} \cdot \frac{126}{330}$

Simplify: $\frac{15}{1365} = \frac{1}{91}, \quad \frac{126}{330} = \frac{21}{55}$

Hence, $P = \frac{1}{91}\cdot \frac{21}{55} = \frac{21}{5005} = \frac{3}{715}$

5. Check the options

The required probability is: $\boxed{\frac{3}{715}}$

So the correct option is: C.