Let the number obtained on the die be $k$.

Since a fair die is rolled once, each of $k=1,2,3,4,5,6$ occurs with probability $\frac{1}{6}$.

If $k$ balls are drawn, the probability that all $k$ are white is $\frac{\binom{6}{k}}{\binom{10}{k}}$ because there are $6$ white balls out of total $10$ balls.

Hence the required probability is $P=\sum_{k=1}^{6} P(\text{die}=k)\cdot P(\text{all white}\mid k)$ $=\frac{1}{6}\sum_{k=1}^{6}\frac{\binom{6}{k}}{\binom{10}{k}}.$

Compute each term:

• For $k=1$: $\frac{\binom{6}{1}}{\binom{10}{1}}=\frac{6}{10}=\frac{3}{5}$

• For $k=2$: $\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{45}=\frac{1}{3}$

• For $k=3$: $\frac{\binom{6}{3}}{\binom{10}{3}}=\frac{20}{120}=\frac{1}{6}$

• For $k=4$: $\frac{\binom{6}{4}}{\binom{10}{4}}=\frac{15}{210}=\frac{1}{14}$

• For $k=5$: $\frac{\binom{6}{5}}{\binom{10}{5}}=\frac{6}{252}=\frac{1}{42}$

• For $k=6$: $\frac{\binom{6}{6}}{\binom{10}{6}}=\frac{1}{210}$

Add them: $\frac{3}{5}+\frac{1}{3}+\frac{1}{6}+\frac{1}{14}+\frac{1}{42}+\frac{1}{210}.$

Taking LCM $210$, $\frac{126+70+35+15+5+1}{210}=\frac{252}{210}=\frac{6}{5}.$

Therefore, $P=\frac{1}{6}\cdot \frac{6}{5}=\frac{1}{5}.$

So the correct option is: $\boxed{\text{C } \frac{1}{5}}$

Checking options systematically:

• A: $\frac{1}{4}$ — incorrect
• B: $\frac{9}{50}$ — incorrect
• C: $\frac{1}{5}$ — correct
• D: $\frac{11}{50}$ — incorrect