When the radius of the earth is reduced by $1\%$ without changing the mass, then change in acceleration due to gravity will be

$\text{g}=\frac{\text{GM}}{{\text{R}}^{\text{2}}}$

${\text{gR}}^{\text{2}}=\text{GM}$

Differentiating w.r.to R on both sides

$\frac{\text{d}}{\text{dR}}\left({\text{gR}}^{\text{2}}\right)=\frac{\text{d}}{\text{dR}}\text{GM}$

$\text{g2R}+\frac{\text{dg}}{\text{dR}}{\text{.R}}^{\text{2}}=0$

$\Rightarrow \Delta {\text{gR}}^{\text{2}}=-\text{g2R}\Delta \text{R}$

$\Rightarrow \frac{\Delta \text{g}}{\text{g}}=\frac{-2\Delta \text{R}}{\text{R}}$

Given that radius of earth is reduced by $1\%$

$\frac{\Delta \text{R}}{\text{R}}\times 100=-1\%\frac{\Delta \text{g}}{\text{g}}\times 100=?$

$\frac{\Delta \text{g}}{\text{g}}\times 100=-2\left(-1\right)\times 100$

$\frac{\Delta \text{g}}{\text{g}}\times 100=2\%$