Given, distance of bottom of hole from the surface of earth (d) = half the radius of earth = $\frac{R_e}{2}$
If g be the value of gravitational acceleration on the surface of earth, then weight of body
$mg=300N$
If g’ be the gravitational acceleration at the bottom of hole, then
$g^{\text{'}}=g\left(1-\frac{d}{R_e}\right)=g\left(1-\frac{\frac{R_e}{2}}{R_e}\right)\Rightarrow g^{\text{'}}=\frac{g}{2}$
$\therefore$ Weight of the body on the bottom of hole,
$m{g}^{\text{'}}=\frac{mg}{2}=\frac{300}{2}=150N$