Solution

Paramagnetism is directly proportional to the number of unpaired electrons. The complex having a number of unpaired electrons will show paramagnetism.
In the above given complexes,
$$A-$$ $$Fe[(CN)_{6}]^{4-}$$, oxidation state of $$Fe$$ is $$+2$$ (i.e., $$x-6=-4$$).Hence, $$Fe^{+2}$$ has $$3d^{6}$$ electrons in the outer most orbit, ($$t_{2g}^{6}$$ and $$e_{g}^{0}$$), so all the electrons are paired, ($$CN$$ is strong field ligand) as shown in diagram ($$n=0$$). So, it will be diamagnetic.
$$B-$$ $$Ni[(CO)_{4}]$$, oxidation state of $$Ni$$ is $$0$$ (i.e., $$x-0=0$$).Hence, $$Ni^{0}$$ has $$3d^{10}$$, so all the electrons are paired ($$e_{g}^{4}$$ and $$t_{2g}^{6}$$) ($$n=0$$). So, it will be diamagnetic.
$$C-$$ $$Ni[(CN)_{4}^{2-}]$$, oxidation state of $$Ni$$ is $$+2$$ (i.e., $$x-4=-2$$).Hence, $$Ni^{+2}$$ has $$3d^{8}$$ electrons in outer most orbit, and this complex has strong field ligand $$(CN)$$, which forces the electron to get paired ($$n=0$$). Hence, diamagnetic.
$$D-$$ $$Co[(F)_{6}]^{3-}$$, oxidation state of $$Co$$ is $$+3$$ (i.e., $$x-6=-3$$).Hence, $$Co^{+3}$$ has $$3d^{6}$$ electrons in the outer most orbit, and ligand $$(F)$$ is a weak field ligand.Hence, four electron will remain unpaired ($$t_{2g}^{4}$$ and $$e_{g}^{2}$$) $$(n=4)$$. So, as explained above, the highest number of unpaired electrons is $$(n=4)$$ for $$Co[(F)_{6}]^{-3}$$.Hence, it shows paramagnetism.