1. Identify the cell and the electrode reaction

The cell is:

$\text{M} \mid \text{M}^{2+}(\text{saturated from } MX_2) \;||\; \text{M}^{2+}(0.001\,\text{mol dm}^{-3}) \mid \text{M}$

For the electrode

$\text{M}^{2+} + 2e^- \rightleftharpoons \text{M}$

this is a concentration cell, so the emf depends only on the difference in $\text{M}^{2+}$ concentration.

2. Use Nernst equation for concentration cell

For a concentration cell with same metal electrodes:

$E = \frac{0.059}{n} \log \frac{[\text{M}^{2+}]_{\text{higher}}}{[\text{M}^{2+}]_{\text{lower}}}$

Here, $n=2$.

Given:

$E = 0.059\,\text{V}$

So,

$0.059 = \frac{0.059}{2} \log \frac{[\text{M}^{2+}]_{\text{higher}}}{[\text{M}^{2+}]_{\text{lower}}}$

Hence,

$1 = \frac{1}{2} \log \frac{[\text{M}^{2+}]_{\text{higher}}}{[\text{M}^{2+}]_{\text{lower}}}$

$\log \frac{[\text{M}^{2+}]_{\text{higher}}}{[\text{M}^{2+}]_{\text{lower}}} = 2$

$\frac{[\text{M}^{2+}]_{\text{higher}}}{[\text{M}^{2+}]_{\text{lower}}} = 10^2 = 100$

One side has

$[\text{M}^{2+}] = 0.001 = 10^{-3}\,\text{mol dm}^{-3}$

The saturated solution of sparingly soluble salt must have the lower concentration. Therefore,

$\frac{10^{-3}}{[\text{M}^{2+}]_{\text{sat}}} = 100$

So,

$[\text{M}^{2+}]_{\text{sat}} = 10^{-5}\,\text{mol dm}^{-3}$

3. Relate solubility to $K_{sp}$

For

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

Let solubility be $s$ mol dm$^{-3}$.

Then,

$[M^{2+}] = s, \qquad [X^-] = 2s$

Given saturated solution has

$[M^{2+}] = s = 10^{-5}$

Thus,

$K_{sp} = [M^{2+}][X^-]^2 = s(2s)^2 = 4s^3$

$K_{sp} = 4(10^{-5})^3 = 4 \times 10^{-15}$

4. Match with options

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

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.