Explanation

(b) Acceleration due to gravity at a height $h$ from earth’s surface
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{'} = \frac{GM}{(R + h)^{2}}$
$Since,\ \ \ \ \ \ \ \ \ \ g^{'} = \frac{g}{100}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{g}{100} = \ \frac{GM}{(R + h)^{2}}$
$or\ \ \ \ \frac{(R + h)^{2}}{100} = \frac{GM}{g}$
$or\ \ \ \ \ \frac{(R + h)^{2}}{100} = R^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack \therefore g = \frac{GM}{R^{2}} \right\rbrack$
$or\ \ \ \ \ \ \ \ \ \ \ R + h = 10R$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = 9R$