To determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that $3^n - 3$ is a multiple of $7$.

Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a = bk$. So in our case, we need to find how many $n$ satisfy the equation $3^n - 3 = 7k$ for some integer $k$.

Notice that $3^n - 3 = 3(3^{n-1} - 1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:

$3^1 \equiv 3 \pmod{7}$

$3^2 \equiv 9 \equiv 2 \pmod{7}$

$3^3 \equiv 27 \equiv 6 \pmod{7}$

$3^4 \equiv 81 \equiv 4 \pmod{7}$

$3^5 \equiv 243 \equiv 5 \pmod{7}$

$3^6 \equiv 729 \equiv 1 \pmod{7}$

We observe that $3^n \pmod{7}$ follows a cycle of length 6. So, $3^{n-1} \pmod{7}$ also follows the same cycle, but shifted:

$3^0 \equiv 1 \pmod{7}$

$3^1 \equiv 3 \pmod{7}$

$3^2 \equiv 2 \pmod{7}$

$3^3 \equiv 6 \pmod{7}$

$3^4 \equiv 4 \pmod{7}$

$3^5 \equiv 5 \pmod{7}$

We want $3(3^{n-1} - 1) \equiv 0 \pmod{7}$, which means that $3^{n-1} - 1 \equiv 0 \pmod{7}$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.

Now let's find the multiples of 6 between 10 and 100:

$12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96$

Adding 1 to each of these values, we get the set of natural numbers $n$ that satisfy the given condition:

$13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97$

There are 15 elements in this set. Therefore, the number of elements in the given set is $\boxed{15}$.