Solution:
We construct an order-preserving bijection between positive multiples of $7$ in base $10$ whose digits are all less than $3$ and positive multiples of $7$. For any multiple of $7$ in base $10$ with digits all less than $3$, interpret it as a base $3$ number and convert it to a base $10$ decimal, which will be a multiple of $7$. For example, $1022$ would map to $1022_3 = 35_{10}$.

We first show that this is a valid mapping. Let $a_n \ldots a_1 a_0$ be an arbitrary multiple of $7$ (in base $10$) whose digits are all less than $3$. This has a natural interpretation in base $3$, and converting this interpretation to base $10$ gives
$$a_n \ldots a_1 a_{0_3} = a_n \cdot 3^n + \cdots + a_1 \cdot 3^1 + a_0 \cdot 3^0 \equiv a_n \cdot 10^n + \cdots + a_1 \cdot 10^1 + a_0 \cdot 10^0 = a_n \ldots a_1 a_0 \pmod{7},$$
which is a multiple of $7$.

This mapping is an injection because base conversion is an injection, and this mapping is also a surjection because for any multiple of $7$ in base $10$, we can convert it to base $3$ and interpret this number as a base $10$ decimal. (Reversing the above steps, we see the resulting decimal number is a multiple of $7$.) Furthermore, for two positive multiples of $7$ whose digits are less than $3$, the larger one will have a larger base $3$ representation, so this mapping is order-preserving.

Thus, this mapping is an order-preserving bijection between multiples of $7$ in base $10$ with digits less than $3$ and multiples of $7$ in base $3$. Therefore, the answer is the preimage of the 100th smallest positive multiple of $7$, which is equal to $700_{10} = \boxed{221221}_{3}$.