The radius of earth is $R$ and acceleration due to gravity on its surface is $g$. The height at which the acceleration due to gravity becomes $\dfrac{g}{8}$ is:

The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula $g' = g \left( \dfrac{R}{R+h} \right)^{2}$, where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.

Given that $g' = \dfrac{g}{8}$, we substitute this into the equation:

$\dfrac{g}{8} = g \left( \dfrac{R}{R+h} \right)^{2}$

Dividing both sides by $g$ gives:

$\dfrac{1}{8} = \left( \dfrac{R}{R+h} \right)^{2}$

Taking the square root of both sides:

$\dfrac{1}{2\sqrt{2}} = \dfrac{R}{R+h}$

Rearranging the terms to solve for $h$:

$R + h = 2\sqrt{2} R$

$h = 2\sqrt{2} R - R$

$h = (2\sqrt{2} - 1) R$

Answer: $(2 \sqrt{2}-1) R$