To solve this problem, we need to determine the van't Hoff factor $i$ for each electrolyte.$\\$The van't Hoff factor $i$ is defined as the number of particles the compound dissociates into when dissolved in water.$\\$For $Al_2(SO_4)_3: \\$• $Al_2(SO_4)_3$ dissociates into 2 $Al^{3+}$ ions and 3 $SO_4^{2-}$ ions.$\\$• Total ions produced = 2 + 3 = 5.$\\$• Therefore, $i = 5. \\$Now, let's calculate $i$ for each option:$\\$Option 1: $K_3[Fe(CN)_6] \\$• $K_3[Fe(CN)_6]$ dissociates into 3 $K^+$ ions and 1 $[Fe(CN)_6]^{3-}$ ion.$\\$• Total ions produced = 3 + 1 = 4.$\\$• Therefore, $i = 4. \\$Option 2: $Al(NO_3)_3 \\$• $Al(NO_3)_3$ dissociates into 1 $Al^{3+}$ ion and 3 $NO_3^-$ ions.$\\$• Total ions produced = 1 + 3 = 4.$\\$• Therefore, $i = 4. \\$Option 3: $K_4[Fe(CN)_6] \\$• $K_4[Fe(CN)_6]$ dissociates into 4 $K^+$ ions and 1 $[Fe(CN)_6]^{4-}$ ion.$\\$• Total ions produced = 4 + 1 = 5.$\\$• Therefore, $i = 5. \\$Option 4: $K_2SO_4 \\$• $K_2SO_4$ dissociates into 2 $K^+$ ions and 1 $SO_4^{2-}$ ion.$\\$• Total ions produced = 2 + 1 = 3.$\\$• Therefore, $i = 3. \\$The electrolyte with the same van't Hoff factor as $Al_2(SO_4)_3$ is $K_4[Fe(CN)_6]$ (Option 3).$\\$Therefore, the correct option is 3.