$$ \begin{aligned} & y=\frac{\mathrm{MgL}^3}{4 \mathrm{bd}^3 \delta} \\ & \frac{\Delta y}{y}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{3 \Delta \mathrm{~L}}{\mathrm{~L}}+\frac{\Delta \mathrm{b}}{\mathrm{~b}}+\frac{3 \Delta \mathrm{~d}}{\mathrm{~d}}+\frac{\Delta \delta}{\delta} \\ & \frac{\Delta y}{y}=\frac{10^{-3}}{2}+\frac{3 \times 10^{-3}}{1}+\frac{10^{-2}}{4}+\frac{3 \times 10^{-2}}{4}+\frac{10^{-2}}{5} \\ & =10^{-3}[0.5+3+2.5+7.5+2]=0.0155 \end{aligned} $$ Option (2)