The velocity of the body just before hitting the ground, due to gravitational acceleration, is given by $$v_{1} = \sqrt{2gh_{1}}$$, and the velocity just after hitting the ground, when it rebounds to a height $$h_{2}$$, is given by $$v_{2} = \sqrt{2gh_{2}}$$.

According to the problem, the ratio $$\frac{v_{1}}{v{2}} = 4$$. Therefore, we can write $$\frac{\sqrt{2gh_{1}}}{\sqrt{2gh_{2}}} = 4$$ or equivalently $$\frac{h_{1}}{h_{2}} = 4^2 = 16$$.

The loss in kinetic energy due to the collision with the ground is given by the difference between the initial kinetic energy $$K_{1} = \frac{1}{2} m v_{1}^2$$ and the final kinetic energy $$K_{2} = \frac{1}{2} m v_{2}^2$$, where m is the mass of the body.

Substituting $$v_{1} = \sqrt{2gh_{1}}$$ and $$v_{2} = \sqrt{2gh_{2}}$$ into these expressions, we get $$K_{1} = mgh_{1}$$ and $$K_{2} = mgh_{2}$$.

The loss in kinetic energy is then $$\Delta K = K_{1} - K_{2} = mgh_{1} - mgh_{2}$$.

The percentage loss in kinetic energy is given by

$$\frac{\Delta K}{K_{1}} \times 100 = \frac{mgh_{1} - mgh_{2}}{mgh_{1}} \times 100 = \frac{h_{1} - h_{2}}{h_{1}} \times 100$$.

Since $$h_{1}/h_{2} = 16$$, we can write $$h_{2} = h_{1}/16$$, so the percentage loss in kinetic energy is

$$\frac{h_{1} - h_{1}/16}{h_{1}} \times 100 = 100(1 - \frac{1}{16}) = 100 \times \frac{15}{16} = \frac{375}{4}$$.

So, the value of $$x$$ is 375.