The value of acceleration due to gravity at height $h$ from earth surface will become half its value on the surface if (R $=$ radius of earth)

The correct option is C. $h=(\sqrt{2}-1)R$

The gravitational force acting on mass m on the surface of Earth is

$F=\frac{GMm}{R^2}$

$mg=\frac{GMm}{R^2}$

$g=\frac{GM}{R^2}$

$g{R}^2=GM$.............$\left(1\right)$
where, $g$ is acceleration due to gravity, $G$ is gravitational constant, $M$ is mass of Earth and R is radius of Earth.
Now, let the value of acceleration due to gravity at height h be g' then we have
$g^{\prime }=\frac{GM}{(R+h{)}^2}$

As per the problem,
$g^{\prime }=\frac{g}{2}=\frac{GM}{(R+h{)}^2}$

$g(R+h{)}^2=2g{R}^2$................from $\left(1\right)$
$(R+h{)}^2=2R^2$
Taking roots from both sides, we get
$R+h=\sqrt{2}R$
$\therefore h=\sqrt{2}R-R$
$\therefore h=(\sqrt{2}-1)R$