If the radius of earth shrinks by 2 % while its mass remains same....

The acceleration due to gravity on the surface of the earth is given by, $g = \dfrac{GM}{{R}^{2}}$.

The radius shrunk by $2$. Therefore, $\dfrac{∆ R}{R} \times 100 = - 2$.

As the mass is constant and the change in the radius of the earth is relatively small, we can write

$\dfrac{∆ g}{g} = - 2\dfrac{∆ R}{R} ⇒ \dfrac{∆ g}{g} \times 100 = 4$

Hence, the acceleration due to gravity increases by $4$.