Explanation

(a) At an altitude $h$ the acceleration due to gravity is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} = g\left( 1 - \frac{2h}{R_{e}} \right)$
$or\ \ \ \ \ \ \ \ \ \ \ \ mg^{'} = mg\left( 1 - \frac{2h}{R_{e}} \right)$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ w^{'} = w\left( 1 - \frac{2h}{R_{e}} \right)$
$\ \ \ \ \ \ \ \ \ \ \frac{99}{100}w = w\left( 1 - \frac{2h}{R_{e}} \right)$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ h = 0.005R_{e}$
At point below the surface of earth at depth $h$. The weight of body given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ w^{'} = w\left( 1 - \frac{2h}{R_{e}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{w^{'}}{w} = 0.995$
$\ \ \ \ \ \ \ \ \%\mathrm{\Delta}w = \frac{(1 - 0.995)w}{w} \times 100$
$\%\mathrm{\Delta}w = 0.5\%(decreases)$