A body weight W , is projected vertically upwards from earth's...

Final distance from the centre of earth wil be $R + 9R = 10R$.

Now acceleration due to gravity at any height(distance from centre r) is given by,

$g = \dfrac{GM}{{r}^{2}}$.

Therefore,

${g}^{'} = \dfrac{GM}{{\left(10R\right)}^{2}}$

Now, value of $g$ at surface will be, $\dfrac{GM}{{R}^{2}}$.

Hence,

${g}^{'} = \left(\dfrac{g}{100}\right)$.

Then, ${W}^{'} = m{g}^{'} = \dfrac{mg}{100} = \dfrac{W}{100}$