Explanation:

We need to find the number of integers $N$ such that $18 \le N \le 108$ and $N$ is a multiple of $3$.\nThis means $N$ can be expressed as $3k$ for some integer $k$.\nSo, we have the inequality: $18 \le 3k \le 108$. \nTo find the range of $k$, we divide all parts of the inequality by $3$: $\frac{18}{3} \le \frac{3k}{3} \le \frac{108}{3}$ $6 \le k \le 36$ \nThis means $k$ can take any integer value from $6$ to $36$, inclusive. To count the number of integers in an inclusive range $[a, b]$, the formula is $b - a + 1$.\nNumber of multiples $= 36 - 6 + 1 = 30 + 1 = 31$. \nLet's evaluate the options:\nOption (1) $29$: This would be $36 - 6 - 1$, which is incorrect.\nOption (2) $30$: This would be $36 - 6$, which excludes one of the endpoints, so it's incorrect.\nOption (3) $31$: This matches our calculated value.\nOption (4) $32$: This would be $36 - 6 + 2$, which is incorrect. \nTherefore, there are $31$ multiples of $3$ between $18$ and $108$, both inclusive. The correct option is (3) 31.