If the radius of a sphere is measured as 9 cm with an error of 0.03 m,

Let x be the radius and y be the surface area of the sphere.

Then,

$x=9$

$\Delta x=0.03 \mathrm{~m}=3 \mathrm{~cm}$

$\Rightarrow x+\Delta x=9+3=12 \mathrm{~cm}$

$y=4 \pi \mathrm{x}^{2}$

For $\mathrm{x}=9$

$\mathrm{y}=4 \pi \times 9^{2}=324 \pi$

$\frac{\mathrm{dy}}{\mathrm{dx}}=8 \pi \mathrm{x}$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=9}=72 \pi$

$\therefore \Delta y=d y=\frac{d y}{d x} d x=72 \pi \times 3=216 \pi \mathrm{cm}^{2}$

Therefore, the approximate error in the surface area is $216 \pi \mathrm{cm}^{2}$.

Disclaimer: This solution has been created according to the question given in the book. However, the solution given in the book is incorrect.