The formula for the time period of a simple pendulum is $T = 2\pi\sqrt{L/g}$, which gives $g = 4\pi^2 \frac{L}{T^2}$. The relative error in $g$ is $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$. Given: $L = 100$ cm, $\Delta L = 1$ mm = $0.1$ cm. Total time $t = 200$ s for $n=100$ oscillations, so $T = t/n$ and $\Delta T/T = \Delta t/t$. Here $\Delta t = 1$ s. Substituting: $\frac{\Delta g}{g} = \frac{0.1}{100} + 2\left(\frac{1}{200}\right) = 0.001 + 0.01 = 0.011$. Expressed as a percentage: $0.011 \times 100 = 1.1\%$. Option 0.1% only considers length, and 2.1% or 0.6% result from incorrect application of the power rule for $T$ or arithmetic errors.