Explanation:
When a body is taken below the surface of the Earth, the gravitational force acting on it increases due to the decreasing distance from the center of the Earth. The weight of the body is given by the formula:

Weight = m*g

where m is the mass of the body and g is the acceleration due to gravity. The value of g changes with the distance from the center of the Earth and is given by:

g = G*M/(r^2)

where G is the universal gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.

Calculation:
Given that the body is taken 64 km below the surface of the Earth. The radius of the Earth is 6400 km.

The distance of the body from the center of the Earth would be:

r = R + h

where R is the radius of the Earth and h is the depth below the surface of the Earth.

r = 6400 + 64 = 6464 km

The acceleration due to gravity at this distance would be:

g = G*M/(r^2)

g = 6.67*10^-11 * 5.97*10^24 / (6464*1000)^2

g = 9.665 m/s^2

The weight of the body at this distance would be:

Weight = m*g

Let the initial weight of the body be W1 and the final weight be W2.

W2 = Weight = m*g = m*9.665

Let the initial distance of the body from the center of the Earth be r1. The acceleration due to gravity at this distance would be:

g1 = G*M/(r1^2)

g1 = 6.67*10^-11 * 5.97*10^24 / (6400*1000)^2

g1 = 9.81 m/s^2

The initial weight of the body would be:

W1 = m*g1 = m*9.81

The percentage decrease in weight would be:

% decrease = (W1 - W2)/W1 * 100

% decrease = (m*9.81 - m*9.665)/(m*9.81) * 100

% decrease = 1.38%

Therefore, the weight of the body decreases by 1.38% when it is taken 64 km below the surface of the Earth.