The acceleration due to gravity on the surface of earth is, \[g=\frac{GM}{R^2}\;\;\text{__(1)}\]

Differentiating both sides, we get, \[dg=-\frac{2GM}{R^3}dR\;\;\text{__(2)}\]

Dividing equation $(2)$ by $(1)$, we get, \[\frac{dg}{g}=-\frac{2dR}{R}\]

Percentage change in $g$ is, \[\frac{dg}{g}×100\%=-2\left(\frac{dR}{R}×100\%\right)\]

The radius of earth is shrunk by $1\%$, \[\therefore\frac{dR}{R}×100\%=-1\%\]

Now, \[\frac{dR}{R}×100\%=(-2)(-1\%)=2\%\]

Hence, the acceleration due to gravity on the surface of earth will increase by 2%.