Explanation

(c) If $M$ be mass of earth and $R$ its radius, the acceleration due to gravity is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Where, $G$ is gravitational constant.
$Given,\ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 0.99R$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} = \frac{GM}{(0.99R)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.02\left( \frac{GM}{R^{2}} \right)$
From Eq. (i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} = 1.02g$
Hence, acceleration due to gravity increases by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} - g = 1.02 - 1 = 0.02g$
Hence, percentage increases =2%.