What happens to the weight of the body if the weight becomes 1/16 at a certain. Also, consider the radius of the earth to be R.

The correct option is 3R.

Let the mass of the body be _m and the mass of the Earth be _M. The radius of the Earth is given as _R.

The weight of a body on the surface of the Earth is given by the gravitational force acting on it:
$$W_1=\frac{GMm}{R^2}$$

Let the weight of the body become $\frac{1}{16}$ of its weight on the Earth's surface at a distance _d from the center of the Earth (where $d=R+h$ and _h is the height above the Earth's surface). The weight at this point is:
$$W_2=\frac{GMm}{d^2}$$

According to the problem, the weight at this distance is $\frac{1}{16}$ of the weight on the surface:
$$W_2=\frac{1}{16}{W}_1$$

Substituting the expressions for $W_1$ and $W_2$:
$$\frac{GMm}{d^2}=\frac{1}{16}\left(\frac{GMm}{R^2}\right)$$

By canceling the common term _GMm from both sides, we get:
$$\frac{1}{d^2}=\frac{1}{16R^2}$$

Taking the reciprocal of both sides:
$$d^2=16R^2$$

Taking the square root on both sides:
$$d=4R$$

Since the distance from the center of the earth is $d=R+h$, where _h is the height above the surface of the earth, we can solve for _h:
$$R+h=4R$$

Solving for _h:
$$h=4R-R=3R$$

Therefore, the height above the surface of the Earth at which the weight becomes 1/16 of its value on the surface is 3R.