If the radius of the earth were shrink by one per cent , its mass remaining the same, the value of ‘g’ on the earth’s surface would

Surface gravity g = GM/R^2, so if Earth's mass M is constant and radius R decreases to R' = 0.99R, the new gravity g' = GM/(0.99R)^2 = g/(0.99^2). Thus g'/g = 1/(0.99^2) ≈ 1.020408, an increase of about 2.04% (rounded to 2%). The inverse-square dependence and the numerical substitution give the result that g increases by ≈2% when radius shrinks by 1% with mass unchanged .