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

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then,

$r=9 \mathrm{~m}$ and $\Delta r=0.03 \mathrm{~m}$

Now, the surface area of the sphere (S) is given by,

$S=4 \pi r^{2}$

$\therefore \frac{d S}{d r}=8 \pi r$

$\therefore d S=\left(\frac{d S}{d r}\right) \Delta r$

$=(8 \pi r) \Delta r$

$=8 \pi(9)(0.03) \mathrm{m}^{2}$

$=2.16 \pi \mathrm{m}^{2}$

Hence, the approximate error in calculating the surface area is $2.16 \pi \mathrm{m}^{2}$.