The atomic number of the element unnilennium is:

We begin by recalling the IUPAC systematic nomenclature that was used for newly discovered elements before they received their permanent names. The rule states that each individual digit of the atomic number is replaced by a specific Latin-Greek prefix and, after writing the three (or more) prefixes in sequence, we add the common ending “ium.” The mapping of digits to prefixes is as follows: $$0 \rightarrow \text{nil}, \; 1 \rightarrow \text{un}, \; 2 \rightarrow \text{bi}, \; 3 \rightarrow \text{tri}, \; 4 \rightarrow \text{quad}, \; 5 \rightarrow \text{pent}, \; 6 \rightarrow \text{hex}, \; 7 \rightarrow \text{sept}, \; 8 \rightarrow \text{oct}, \; 9 \rightarrow \text{enn}.$$

Now we examine the given temporary name “unnilennium.” We separate it into its component prefixes:

“un” + “nil” + “enn” + “ium.”

Using the digit-prefix chart stated above, we translate each prefix back into its numerical value:

$$\text{un} \rightarrow 1, \quad \text{nil} \rightarrow 0, \quad \text{enn} \rightarrow 9.$$

Placing these digits in the same order in which their prefixes appeared, we form the atomic number:

$$1\,0\,9 \;=\; 109.$$

So, the element whose temporary systematic name was “unnilennium” possesses the atomic number $$109$$.

Looking at the options given, we see that $$109$$ corresponds to Option A.

Hence, the correct answer is Option A.