1. We are given the standard reduction potentials:

$Sn^{2+} + 2e^- \to Sn, \qquad E_1^\circ = -0.140\,\text{V}$

$Sn^{4+} + 4e^- \to Sn, \qquad E_2^\circ = +0.010\,\text{V}$

We need to find:

$Sn^{4+} + 2e^- \to Sn^{2+}, \qquad E^\circ_{Sn^{4+}/Sn^{2+}} = ?$

2. Standard electrode potentials cannot be added directly, so we use the relation with standard Gibbs free energy:

$\Delta G^\circ = -nFE^\circ$

3. Write the target reaction as a combination of the given reactions.

Reaction (A): $Sn^{2+} + 2e^- \to Sn \qquad (E_A^\circ = -0.140\,\text{V})$

Reverse it: $Sn \to Sn^{2+} + 2e^-$ Then $\Delta G_A^{\circ,\text{rev}} = -\Delta G_A^\circ$

Reaction (B): $Sn^{4+} + 4e^- \to Sn \qquad (E_B^\circ = +0.010\,\text{V})$

Add reaction (B) to reverse of (A) multiplied suitably:

Actually, $Sn^{4+} + 4e^- \to Sn$ $Sn \to Sn^{2+} + 2e^-$

$Sn^{4+} + 2e^- \to Sn^{2+}$

So Gibbs energies add:

4. Compute $\Delta G^\circ$ values.

For reaction (B): $\Delta G_B^\circ = -nFE_B^\circ = -4F(0.010) = -0.040F$

For reaction (A): $\Delta G_A^\circ = -2F(-0.140) = +0.280F$

So for reverse of (A): $\Delta G_A^{\circ,\text{rev}} = -0.280F$

Hence for $Sn^{4+} + 2e^- \to Sn^{2+}$

$\Delta G^\circ = \Delta G_B^\circ + \Delta G_A^{\circ,\text{rev}}$ $= -0.040F + (-0.280F)$ $= -0.320F$

5. Now convert back to electrode potential for the 2-electron reaction:

$\Delta G^\circ = -nFE^\circ$ $-0.320F = -2F E^\circ$

$E^\circ = \frac{0.320}{2} = 0.160\,\text{V}$

6. The question asks for the magnitude in the form:

$\underline{\hspace{2cm}} \times 10^{-2}\,\text{V}$

Since $0.160\,\text{V} = 16 \times 10^{-2}\,\text{V}$

So the required nearest integer is:

$\boxed{16}$