The weight of a body at the centre of earth is

We know that, on going below the earth surface, $g$ decreases, which is given by

$g^{\prime }=g\left(1-\frac{d}{R_e}\right)$ .....(i)

where, $d=$ depth

$R_e=$ Radius of earth.

At the centre of earth, $d=R_e$ [ $\therefore$ from Equation (i)]

$g^{\prime }=g\left(1-\frac{R_e}{R_e}\right)$

$g^{\prime }=0$

Hence, weight of the body at the centre,

$w=m{g}^{\prime }$

$=m\times 0$

$=0$