In the sixth period, the orbitals that are filled are:

First, let us recall the Aufbau principle, which states that electrons occupy the available orbitals in the order of increasing $$(n+l)$$ value, and if two orbitals have the same $$(n+l)$$ value, then the one with the lower $$n$$ (principal quantum number) is filled first.

Now, we list all the orbitals in the increasing order of $$(n+l)$$ up to the region where the sixth period ends (i.e., up to the noble-gas element radon, $$Z = 86$$):

$$1s,\, 2s,\, 2p,\, 3s$$, $$3p,\, 4s,\, 3d,\, 4p$$, $$5s,\, 4d,\, 5p,\, 6s$$, $$4f,\, 5d,\, 6p,\, 7s,\ldots$$

We see that the very first orbital whose principal quantum number is $$6$$ is $$6s$$. So, after finishing $$5p$$ (which completes the fifth period at xenon, $$Z = 54$$), the next orbital to receive electrons is $$6s$$.

After filling the $$6s$$ orbital with two electrons, we check the next entry in our ordered list of orbitals. That orbital is $$4f$$. Although its principal quantum number $$n = 4$$ is numerically less than 6, its $$(n+l)$$ value $$=4+3=7$$ is larger than that of $$6s$$ $$\big(n+l = 6+0=6\big)$$, so, according to the Aufbau ordering, $$4f$$ comes directly after $$6s$$. Hence, as we progress through the sixth period (elements $$Z = 57$$ to $$Z = 70$$, the lanthanoids), the $$4f$$ subshell gets filled.

Once $$4f$$ is completely filled with its $$14$$ electrons, the next orbital in the list is $$5d$$. Thus, elements $$Z = 71$$ to $$Z = 80$$ (the sixth-period transition elements) add electrons to $$5d$$.

When $$5d$$ reaches its capacity of $$10$$ electrons, the sequence finally moves on to $$6p$$. The $$6p$$ orbitals are filled across the last part of the sixth period, ending with radon, $$Z = 86$$.

Putting all these steps together, the orbitals that receive electrons in the sixth period are, in chronological order, $$6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$$.

Comparing this list with the supplied options, we observe that Option A lists the orbitals exactly as $$6s,\,4f,\,5d,\,6p$$, which matches our derived sequence.

Hence, the correct answer is Option A.