If the radius of the Earth shrinks by 1% while its mass remains unchanged, the acceleration due to gravity on its surface will approximately:

The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Taking natural logarithm and differentiating: $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -1\% = -0.01$, we get $\frac{\Delta g}{g} = -2(-1\%) = +2\%$. Thus, $g$ increases by approximately $2\%$.