First, let us determine the $d$-electron count of the cobalt center in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$.

1. Oxidation state and $d$-electron count

Neutral cobalt (Co) has an atomic number of 27 and an electronic configuration:

$ \mathrm{Co} \; \bigl[\mathrm{Ar}\bigr]\, 3d^7\, 4s^2. $

In the +3 oxidation state, cobalt has lost a total of 3 electrons:

First two electrons are removed from the $4s$ orbital.

The third electron is removed from the $3d$ orbitals.

Thus, $\mathrm{Co}^{3+}$ has

$ 3d^{7-1} = 3d^6. $

So in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$, the cobalt center is $d^6$.

2. High-spin vs Low-spin for $\mathrm{Co}^{3+}$

Even though $\mathrm{H}_2\mathrm{O}$ is generally considered a weak field ligand, the $\mathrm{Co}^{3+}$ ion has a relatively high charge (+3). A higher charge on the metal center usually increases the ligand-field splitting ($\Delta_\mathrm{o}$) significantly compared to lower oxidation states of the same metal.

In most typical octahedral complexes of $\mathrm{Co}^{3+}$, the splitting $\Delta_\mathrm{o}$ is large enough that the complex ends up being low spin.

Electronic configuration in an octahedral field

For a $d^6$ octahedral low-spin complex, all six electrons pair up in the lower-energy $\mathrm{t_{2g}}$ orbitals:

$ \mathrm{t_{2g}^6}\, \mathrm{e_g^0} $

That leaves 0 unpaired electrons.

3. Final answer

Therefore, the number of unpaired $d$-electrons in

$\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$ is $ \boxed{0}. $

Correct Option: A (0)