If the mass of the earth is increased by 1%, density is decreased by 2 % the change in acceleration due to gravity at the surface of earth is

-1%

$g=\frac{G M}{r^2}$  Since the average density of earth $\rho=\frac{M}{(4 / 3) \pi r^3} \Rightarrow r=\left(\frac{3 M}{4 \pi \rho}\right)^{1 / 3}$

$\Rightarrow g=\frac{G M}{\left(\frac{3 M}{4 \pi \rho}\right)^{2 / 3}}=G^{1 / 3}\left(\frac{4 \pi}{3}\right)^{2 / 3} \rho^{2 / 3}$

$\Rightarrow \ln g=\ln \left\{G\left(\frac{4 \pi}{3}\right)^{2 / 3}\right\}+\frac{1}{3} \ln M+\frac{2}{3} \ln \rho$

$\Rightarrow \frac{\delta g}{g}=\frac{\delta M}{M}+\frac{2}{3} \frac{\delta \rho}{\rho} \quad \Rightarrow \frac{\delta g}{g}=\frac{1}{3}(+1 \%)+\frac{2}{3}(-2)%$

⇒ -1%