The acceleration due to gravity at a height ( 1 / 20 ) t h of the radius of earth above the earth's…

The variation in acceleration due to gravity due to height above the earth surface is, $g\text{'}=g\left(1-\frac{2h}{R}\right)$

Given that $g\text{'}=9 \mathrm{m} {\mathrm{s}}^{-2}$and $h=\frac{1}{20}R$therefore, 

$$\begin{gathered} \Rightarrow 9=g\left(1-\frac{1}{10}\right) \\ \Rightarrow g=10 \mathrm{m} {\mathrm{s}}^{-2} \end{gathered}$$

The variation in acceleration due to gravity due to height below the earth surface is $g\text{'}\text{'}=g\left(1-\frac{h}{R}\right)$Here $h=\frac{R}{20}$ therefore, 

$g\text{'}\text{'}=10\left(1-\frac{1}{20}\right) =\frac{19}{2}=9.5 \mathrm{m} {\mathrm{s}}^{-2}$