Let $r$ be the radius of sphere & $∆r$ be the error in measuring the radius.

Then $r=9 \& ∆r=0.03$

Now, the surface area of sphere is given by, $S=4\pi r^2$

$\Rightarrow ∆S=8\pi r∆r$

$\Rightarrow ∆S=8\pi \times 9\times 0.03=2.16\pi {\mathrm{cm}}^2$