${\left[\mathrm{Cr}(\mathrm{CN}{)}_6\right]}^{3-}$ ion, oxidation state of $\mathrm{Cr}$ is +3 and its valence shell electronic configuration is $3{\mathrm{d}}^3$. There are $3$ unpaired electrons in $3\mathrm{d}$ orbital.  

(A). ${\left[\mathrm{Cr}(\mathrm{CN}{)}_6\right]}^{3-}$

No. of unpaired electrons $=3$

 

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ ion, oxidation state of $\mathrm{Fe}$ is +2 and its valence shell electronic configuration is $3{\mathrm{d}}^6$. There are 4 unpaired electrons in 3d orbital.  So, you can say the hybridisation here would be ${\mathrm{sp}}^3{\mathrm{d}}^2$.

(B). ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$

No. of unpaired electrons $=4$

${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ in this oxidation state of central metal atom is $+3$ and it has no unpaired electrons.
(C). ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$

No. of unpaired electrons $=0$

${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$ the oxidation state of $\mathrm{Ni}$ is ${\mathrm{Ni}}^{2+}$ and it has two unpaired electrons.
(D) ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$

No. of unpaired electrons $=2$
So the correct option among the given is $\mathrm{A}$.