In an experiment to find g with a simple pendulum, the length of the thread is measured as 96.0 cm with a possible error of 0.4 cm and the diameter…

The effective length of the pendulum is the thread length plus the bob radius, $L=l+\dfrac d2=96.0+4.0=100.0\ \mathrm{cm}$. Its maximum error is the sum of the absolute errors of the contributions: $\Delta L=\Delta l+\dfrac{\Delta d}{2}=0.4+0.1=0.5\ \mathrm{cm}$, so $\dfrac{\Delta L}{L}=0.5\%$. The period is $T=40.0/20=2.00\ \mathrm{s}$ with $\Delta T=0.5/20=0.025\ \mathrm{s}$, i.e. a fractional error $\dfrac{\Delta T}{T}=\dfrac{0.5}{40.0}=1.25\%$. Since $g\propto L\,T^{-2}$, the maximum fractional error is $\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}=0.5\%+2\times1.25\%=3.0\%$. The answer is therefore $3$. Note that the error in the time is doubled because $T$ enters squared, and the errors are always added, never subtracted, when the maximum error is asked.