Given, $g = \frac{4\pi^2 L}{T^2}$ $L = 10 \text{ cm}$, $\Delta L = 0.1 \text{ cm}$ $T = 100 \text{ s}$, $\Delta T = 1 \text{ s}$

The formula for the relative error in $g$ is: $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$

To find the percentage error, we multiply by 100: $\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \right) \times 100$ $\frac{\Delta g}{g} \times 100 = \left( \frac{0.1}{10} + 2 \times \frac{1}{100} \right) \times 100$ $\frac{\Delta g}{g} \times 100 = \left( \frac{1}{100} + \frac{2}{100} \right) \times 100$ $\frac{\Delta g}{g} \times 100 = \left( \frac{3}{100} \right) \times 100 = 3\%$

Therefore, the percentage error in the measurement of $g$ is $3\%$.