If the percentage error in measuring the surface area of a sphere is $\alpha \%$, then the error in its volume, is

$\frac{3}{2} \alpha \%$

Let r be the radius, S the surface area and V the volume of the sphere. Then,

$S=4 \pi r^2$ and $V=\frac{4}{3} \pi r^3$

Let $\Delta r, \Delta S$ and $\Delta V$ be the errors in r, S and V respectively. Then,

$\Delta S =\frac{d S}{d r} \Delta r$ and $\Delta V=\frac{d V}{d r} \Delta r$

$\Rightarrow \Delta S =8 \pi r \Delta r$ and $\Delta V=4 \pi r^2 \Delta r$

$\Rightarrow \frac{\Delta S}{S}=\frac{8 \pi r}{4 \pi r^2} \Delta r$ and $\frac{\Delta V}{V}=\frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3}$

$\Rightarrow \frac{\Delta S}{S} \times 100=2\left(\frac{\Delta r}{r} \times 100\right)$ and $\frac{\Delta V}{V} \times 100=3\left(\frac{\Delta r}{r} \times 100\right)$

$\Rightarrow \alpha=2\left(\frac{\Delta r}{r} \times 100\right)$ and $\frac{\Delta V}{V} \times 100=3\left(\frac{\Delta r}{r} \times 100\right)$            $\left[∵ \frac{\Delta S}{S} \times 100=\alpha\right.$ (given)$]$

$\Rightarrow \frac{\Delta V}{V} \times 100=3\left(\frac{\alpha}{2}\right)=\frac{3}{2} \alpha$