Explanation:
B Given that, $\mathrm{g}^{\prime}=1 \% \mathrm{~g}$
$\mathrm{g}^{\prime}=\frac{\mathrm{g}}{100}$
Acceleration due to gravity at earth surface,
$g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$
$\frac{g^{\prime}}{g}=\left(\frac{R}{R+h}\right)^{2}$
$\frac{g}{\frac{100}{g}}=\left(\frac{R}{R+h}\right)^{2}$
$\frac{1}{100}=\left(\frac{R}{R+h}\right)^{2}$
$\frac{R}{R+h}=\sqrt{\frac{1}{100}}$
$\frac{R}{R+h}=\frac{1}{10}$
$R+h=10 R$
$h=10 R-R$
$h=9 R$
So, height, $h=9 R$, at which acceleration due to gravity becomes $1 \%$ of its value on the surface.