Osmotic pressure is a colligative property given by the formula $\pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature . For two solutions to have the same osmotic pressure at a given temperature, their effective concentrations ($i \times C$) must be equal. Assuming complete dissociation:

1. 0.1 M $BaCl_2$ and 0.2 M $K_2SO_4$: For $BaCl_2$, $i = 3$ ($Ba^{2+}, 2Cl^-$), so $i \times C = 3 \times 0.1 = 0.3$ M. For $K_2SO_4$, $i = 3$ ($2K^+, SO_4^{2-}$), so $i \times C = 3 \times 0.2 = 0.6$ M. (Not equal)
2. 0.1 M $Na_3PO_4$ and 0.1 M $K_2SO_4$: For $Na_3PO_4$, $i = 4$ ($3Na^+, PO_4^{3-}$), so $i \times C = 4 \times 0.1 = 0.4$ M. For $K_2SO_4$, $i \times C = 3 \times 0.1 = 0.3$ M. (Not equal)
3. 0.2 M NaCl and 0.1 M $K_2SO_4$: For NaCl, $i = 2$ ($Na^+, Cl^-$), so $i \times C = 2 \times 0.2 = 0.4$ M. For $K_2SO_4$, $i \times C = 3 \times 0.1 = 0.3$ M. (Not equal)
4. 0.2 M NaCl and 0.1 M $K_3[Fe(CN)_6]$: For NaCl, $i = 2$, so $i \times C = 2 \times 0.2 = 0.4$ M. For $K_3[Fe(CN)_6]$, $i = 4$ ($3K^+, [Fe(CN)_6]^{3-}$), so $i \times C = 4 \times 0.1 = 0.4$ M. (Equal)

Therefore, 0.2 M NaCl and 0.1 M $K_3[Fe(CN)_6]$ will have the same value of osmotic pressure.