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A 1:3 electrolyte is a compound that dissociates in an aqueous solution to produce one cation and three anions (or vice versa), resulting in a total of four ions. The general formula for such an electrolyte is $[M]X_{3}$.
We examine the dissociation of each coordination complex:
(A) $[Co(NH_{3})_{3}(NO_{2})_{3}]$: This is a non-electrolyte as it does not ionize in water.
(B) $[CoCl_{2}(NH_{3})_{4}]Cl$: Dissociates as $[CoCl_{2}(NH_{3})_{4}]^{+} + Cl^{-}$. This is a 1:1 electrolyte.
(C) $[CoCl(NH_{3})_{5}]Cl_{2}$: Dissociates as $[CoCl(NH_{3})_{5}]^{2+} + 2Cl^{-}$. This is a 1:2 electrolyte.
(D) $[Co(NH_{3})_{6}]Cl_{3}$: Dissociates as $[Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$. This produces one complex cation and three chloride anions.
Since option (D) yields four ions in total (one complex ion and three chloride ions), it represents a 1:3 electrolyte.
Final Answer: [Co(NH_{3})_{6}]Cl_{3}