The weight of an object at a distance d from the center of the Earth is given by:

$$W' = W \left(\frac{R}{d}\right)^2$$

where:

• W' is the weight at distance d,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• d is the distance from the center of the Earth.

In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:

$$\Delta W = \frac{W - W'}{W} \times 100\%$$

where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:

$$\Delta W = \left(1 - \left(\frac{R}{d}\right)^2\right) \times 100\%$$

The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\frac{5}{4}R$. Substituting these values into the equation gives:

$$\Delta W = \left(1 - \left(\frac{6400~km}{\frac{5}{4} \cdot 6400~km}\right)^2\right) \times 100\% = \left(1 - \left(\frac{4}{5}\right)^2\right) \times 100\% = 36\% $$

Therefore, the percentage decrease in the weight of the object when taken to a height of $\frac{1}{4}R$ above the surface of the Earth is 36%.