The relative error in the determination of the surface area of a sphere is $$\alpha$$. Then the relative error in the determination of its volume is:

For a sphere of radius $$r$$, the surface area is given by the well-known formula $$S = 4\pi r^{2}.$$

The relative error in a calculated quantity is defined as $$\dfrac{\Delta Q}{Q},$$ where $$\Delta Q$$ is the absolute (small) error in that quantity $$Q.$$

We are told that the relative error in surface area is $$\alpha,$$ so we may write

$$\frac{\Delta S}{S} = \alpha.$$

To connect this with the radius, we first differentiate the formula for surface area. Taking differentials we have

$$dS = \frac{d}{dr}(4\pi r^{2}) \, dr = 8\pi r \, dr.$$

Dividing by $$S = 4\pi r^{2},$$ we obtain

$$\frac{dS}{S} = \frac{8\pi r \, dr}{4\pi r^{2}} = 2\,\frac{dr}{r}.$$

Replacing the differentials with small errors ($$dS \to \Delta S$$ and $$dr \to \Delta r$$), this becomes

$$\frac{\Delta S}{S} = 2\,\frac{\Delta r}{r}.$$

But $$\dfrac{\Delta S}{S} = \alpha,$$ hence

$$2\,\frac{\Delta r}{r} = \alpha \quad\Longrightarrow\quad \frac{\Delta r}{r} = \frac{\alpha}{2}.$$

Now we turn to the volume, whose formula is $$V = \dfrac{4}{3}\pi r^{3}.$$

Differentiating,

$$dV = \frac{d}{dr}\!\left(\frac{4}{3}\pi r^{3}\right) dr = 4\pi r^{2} \, dr.$$

Dividing by $$V = \dfrac{4}{3}\pi r^{3},$$ we get

$$\frac{dV}{V} = \frac{4\pi r^{2}\, dr}{\dfrac{4}{3}\pi r^{3}} = 3\,\frac{dr}{r}.$$

Substituting $$\dfrac{dr}{r} = \dfrac{\alpha}{2},$$ we have

$$\frac{\Delta V}{V} = 3 \left(\frac{\alpha}{2}\right) = \frac{3}{2}\,\alpha.$$

Therefore, the relative error in the volume is $$\dfrac{3}{2}\alpha.$$

Hence, the correct answer is Option 3.