The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.

If ( $d = \frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \frac{R}{2} = \frac{R}{2}$ ).

Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \frac{R}{2} $) is half its weight at the surface of the Earth.

So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \frac{R}{2} $) is half of that, or 100 N.

Therefore, the correct answer is 100 N.