1. Given data

   The radius of the sphere is measured as $r = (7.50 \pm 0.85)\text{ cm}$

   So,

   • Absolute error in radius = $\Delta r = 0.85\text{ cm}$
   • Measured radius = $r = 7.50\text{ cm}$

2. Volume of a sphere

   The volume is $V = \frac{4}{3}\pi r^3$

3. Error propagation rule

   For a quantity of the form $V \propto r^3$, the fractional error is $\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$

   Therefore, percentage error in volume is $x = 3\left(\frac{\Delta r}{r}\right)\times 100$

4. Substitute the values

   $x = 3\left(\frac{0.85}{7.50}\right)\times 100$

   First compute: $\frac{0.85}{7.50} = 0.1133\ldots$

   Then, $x = 3 \times 0.1133\ldots \times 100$ $x = 0.34 \times 100$ $x = 34.0\%$

5. Nearest integer

   $x \approx 34$

Final Answer

The percentage error in volume is $\boxed{34}$