To find the solubility product (K_sp) for the sparingly soluble salt, MX₂, in a concentration cell setup, we first analyze how the emf is related to the concentration differences across the cell. The reaction at each electrode involves the metal ion M²⁺ and the metal M, whereas the net cell reaction has no change in number of moles on both sides of the equation due to the symmetry of the cell. Thus, the notation for the cell reaction is:

$$ M(s) \longleftrightarrow M^{2+}(aq) + 2e^{-} $$

Given that it's a concentration cell, the emf generated is due to the concentration difference of the M²⁺ ions at the two electrodes. The emf of the cell can be calculated by the Nernst equation:

$$ E = E^{\circ} - \frac{RT}{nF} \ln\left(\frac{[M^{2+}]_{\text{cathode}}}{[M^{2+}]_{\text{anode}}}\right) $$

Since it's a concentration cell, $ E^\circ = 0 $. At 298 K, substituting from the provided conversion (2.303 $ \times $ R $ \times $ 298/F = 0.059 V) and given n = 2 (because two electrons are transferred per metal ion):

$$ E = - \frac{(0.059)}{2} \log\left(\frac{[M^{2+}]_{saturated}}{[M^{2+}]_{0.001 \text{ M}}}\right) $$

Given $ E = 0.059 $ V, we can solve for the concentration of $ M^{2+} $ in the saturated solution:

$$ 0.059 = -0.0295 \log\left(\frac{[M^{2+}]_{saturated}}{0.001}\right) $$

$$ \log\left(\frac{[M^{2+}]_{saturated}}{0.001}\right) = -2 $$

$$ [M^{2+}]_{saturated} = 0.001 \times 10^{-2} $$

$$ [M^{2+}]_{saturated} = 0.00001 \text{ M} $$

The solubility product, K_sp, of MX₂ can now be computed. Let s be the solubility of MX₂ in mol/L. The dissolution of MX₂ is given by:

$$ MX_2 \rightleftharpoons M^{2+} + 2 X^- $$

$$ \quad s \quad\quad\quad s \quad \quad \, 2s $$

The solubility product expression is:

$$ K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 $$

$$ K_{sp} = 4s^3 $$

Since $ s = [M^{2+}]_{saturated} = 0.00001 $ M,

$$ K_{sp} = 4(0.00001)^3 $$

$$ K_{sp} = 4 \times 10^{-15} $$

Thus, the solubility product of MX₂ at 298 K is $ 4 \times 10^{-15} $ mol³ dm^-9, which corresponds to Option B.