If the number of five digit numbers with distinct digits and 2 at the $$10^{th}$$ place is $$336k$$, then $$k$$ is equal to:

We have to count the five-digit natural numbers in which every digit is different and the digit at the tens place (that is, the $$10^{1}$$ place) is a fixed 2.

Let us write a general five-digit number as

$$\overline{d_5\,d_4\,d_3\,d_2\,d_1},$$

where $$d_5$$ is the ten-thousands digit, $$d_4$$ the thousands digit, $$d_3$$ the hundreds digit, $$d_2$$ the tens digit and $$d_1$$ the units digit. According to the statement $$d_2=2$$. All the remaining digits have to be chosen from the other ten numerals $$0,1,3,4,5,6,7,8,9$$ with the restriction that no digit repeats.

Step 1. Choosing $$d_5$$. For a five-digit number the leading digit cannot be $$0$$, and it also cannot be $$2$$ (because $$2$$ is already sitting at the tens place). Hence $$d_5$$ can be chosen from the set $$\{1,3,4,5,6,7,8,9\}$$, which gives

$$8\text{ ways.}$$

Step 2. Choosing $$d_4$$. After we have fixed $$d_5$$ and $$d_2$$, two different digits are already used. Eight digits are still unused: $$0$$ together with the seven digits that are not $$2$$ and not equal to the chosen $$d_5$$. Therefore

$$d_4$$ can be filled in $$8\text{ ways.}$$

Step 3. Choosing $$d_3$$. Now three different digits have been occupied, so

$$d_3$$ can be selected in $$7\text{ ways.}$$

Step 4. Choosing $$d_1$$. With four distinct digits already fixed, six digits remain unused, and any of them may be taken for the units place. Hence

$$d_1$$ can be assigned in $$6\text{ ways.}$$

Total count. By the multiplication principle, the total number of admissible five-digit numbers is

$$8 \times 8 \times 7 \times 6 \;=\;2688.$$

This number is expressed in the statement as $$336k$$, so

$$336k \;=\;2688 \;\;\Longrightarrow\;\; k \;=\;\dfrac{2688}{336} \;=\;8.$$

Hence, the correct answer is Option D.