Calculating the ionic strength for an aqueous solution containing potassium ferricyanide and sodium sulfate
In aqueous solutions, $\ce{K3Fe(CN)6}$ and $\ce{Na2SO4}$ will completely dissociate:
$$\ce{K3Fe(CN)6 (aq) -&gt; 3 K+ (aq) + Fe(CN)6^3- (aq)} \tag1$$
$$\ce{Na2SO4 (aq) -&gt; 2 Na+ (aq) + SO4^2- (aq)} \tag2$$
Thus, from the equation $(1)$, $c_\ce{K+} = 3c_\ce{Fe(CN)6^3-} = 3c_\ce{K3Fe(CN)6}$ and from the equation $(2)$, $c_\ce{Na+} = 2c_\ce{SO4^2-} = 2c_\ce{Na2SO4}$
Hence for condition A $([\ce{Na2SO4}] = 0)$:
$$I_{\mathrm{A}} = \frac {1}{2} \left((1)^2 \cdot (3\cdot10^{-4}) + (-3)^2\cdot(1\cdot10^{-4}) + (1)^2 \cdot 0 + (-2)^2\cdot 0 \right) = 6\times 10^{-4} = 0.6\times 10^{-3}$$
Similarly, for condition B $([\ce{Na2SO4}] = 1\cdot10^{-3})$:
$$I_{\mathrm{B}} = \frac {1}{2} \left((1)^2\cdot(3\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(2\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) \right) = 3.6 \times 10^{-3}$$
And, for condition C $([\ce{Na2SO4}] = 3\cdot10^{-3})$:
$$I_{\mathrm{C}} = \frac {1}{2} \left((3)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(6\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) \right) = 9.6 \times  10^{-3}$$
So on for other conditions. Note that for all calculations, ionic strength of $\ce{K3Fe(CN)6}$ is always $0.6\times 10^{-3}$ since $[\ce{K3Fe(CN)6}] = \pu{1.0 \times  10^{-4} M}$ for all conditions.