Six distinct integers are picked at random from $\{1,2,3,\ldots,10\}$.  What is the probability that, among those selected, the second smallest is $3$?

$ \textbf{(A)}\ \frac{1}{60}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \text{none of these} $
1. **Determine the total number of ways to pick 6 distinct integers from the set $\{1, 2, 3, \ldots, 10\}$:**
   \[
   \binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = 210
   \]

2. **Identify the condition that the second smallest number among the selected integers is 3:**
   - The smallest number must be either 1 or 2.
   - The second smallest number is fixed as 3.
   - We need to choose 4 more numbers from the remaining set $\{4, 5, 6, 7, 8, 9, 10\}$.

3. **Calculate the number of ways to choose the smallest number:**
   - There are 2 choices for the smallest number (either 1 or 2).

4. **Calculate the number of ways to choose the remaining 4 numbers from the set $\{4, 5, 6, 7, 8, 9, 10\}$:**
   \[
   \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35
   \]

5. **Combine the number of ways to choose the smallest number and the remaining 4 numbers:**
   - There are 2 ways to choose the smallest number.
   - There are 35 ways to choose the remaining 4 numbers.
   - Therefore, the total number of favorable outcomes is:
     \[
     2 \times 35 = 70
     \]

6. **Calculate the probability:**
   \[
   \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{70}{210} = \frac{1}{3}
   \]

The final answer is $\boxed{\frac{1}{3}}$