The total number of unpaired electrons present in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]…

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2$
${\mathrm{Co}}^{2+}:\left[\mathrm{Ar}\right] 3{\mathrm{d}}^{7}4{ \mathrm{s}}^{0}4{\mathrm{p}}^0$
For strong field ligand and ${\mathrm{d}}^7$ configuration: ${\mathrm{t}}_{2\mathrm{g}}^{2,2,2} {\mathrm{e}}_{\mathrm{g}}^{1,0}$ 
Total $1$ unpaired electron is present.
$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$
${\mathrm{Co}}^{3+}:\left[\mathrm{Ar}\right]3{ \mathrm{d}}^{6}4{ \mathrm{s}}^{0}4{\mathrm{p}}^0$
The hybridisation of the complex is ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridisation.
${\mathrm{NH}}_3$ acts as strong field ligand, because ${\mathrm{\Delta }}_0>\mathrm{P}.\mathrm{E}.$.
So here all electrons becomes paired.