Answer: A
Let the original radius of the sphere be \(r\) units. After a 3% increase, the new radius becomes: \(r + \frac{3r}{100} = \frac{103r}{100}\) units \(\therefore\) The percentage increase in surface area is: \[ = \frac{4\pi \left(\frac{103r}{100}\right)^2 - 4\pi r^2}{4\pi r^2} \times 100\% = \left[\left(\frac{103}{100}\right)^2 - 1\right] \times 100\% = \frac{10609 - 10000}{10000} \times 100\% = \frac{609}{100}\% = 6.09\% \]
Let the original radius of the sphere be \(r\) units. After a 3% increase, the new radius becomes: \(r + \frac{3r}{100} = \frac{103r}{100}\) units \(\therefore\) The percentage increase in surface area is: \[ = \frac{4\pi \left(\frac{103r}{100}\right)^2 - 4\pi r^2}{4\pi r^2} \times 100\% = \left[\left(\frac{103}{100}\right)^2 - 1\right] \times 100\% = \frac{10609 - 10000}{10000} \times 100\% = \frac{609}{100}\% = 6.09\% \]