Two particles of the same mass m are moving in circular orbits because of force, given by $$F(r) = -\frac{16}{r} - r^3$$. The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:

We are told that both particles are of the same mass $$m$$ and are moving in circular orbits under a central force

$$F(r)= -\frac{16}{r}-r^{3}.$$

The negative sign merely shows that the force is attractive, so for the purpose of balancing the centripetal requirement we will use its magnitude

$$|F(r)|=\frac{16}{r}+r^{3}.$$

For a body of mass $$m$$ in uniform circular motion of radius $$r$$ and speed $$v$$, the centripetal force is given by the standard relation

$$F_{\text{centripetal}}=\frac{m v^{2}}{r}.$$

Because the circular motion is maintained by the given force, we equate the magnitudes:

$$\frac{m v^{2}}{r}= \frac{16}{r}+r^{3}.$$

Multiplying both sides by $$r$$ to clear the denominator, we get

$$m v^{2}= 16 + r^{4}.$$

Now we solve for $$v^{2}$$:

$$v^{2}= \frac{16 + r^{4}}{m}.$$

The kinetic energy of a particle is expressed by the familiar formula

$$K=\frac{1}{2} m v^{2}.$$

Substituting the value of $$v^{2}$$ obtained above, we obtain

$$K=\frac{1}{2} m\left(\frac{16 + r^{4}}{m}\right)=\frac{1}{2}(16 + r^{4}).$$

Notice that the mass $$m$$ cancels out, so the kinetic energy depends only on the radius:

$$K(r)=8 + \frac{r^{4}}{2}.$$

For the first particle, the radius is $$r_{1}=1$$. Substituting $$r=1$$ in the expression for $$K(r)$$, we have

$$K_{1}=8 + \frac{1^{4}}{2}=8 + \frac{1}{2}=8.5.$$

For the second particle, the radius is $$r_{2}=4$$. Substituting $$r=4$$, we get

$$K_{2}=8 + \frac{4^{4}}{2}=8 + \frac{256}{2}=8 + 128=136.$$

We need the ratio of the kinetic energies of the first particle to the second particle:

$$\frac{K_{1}}{K_{2}}=\frac{8.5}{136}.$$

Dividing numerator and denominator by 136, or using a calculator, gives

$$\frac{K_{1}}{K_{2}}\approx 0.0625=6.25 \times 10^{-2}.$$

Among the options provided, $$6 \times 10^{-2}$$ is closest to this numerical value.

Hence, the correct answer is Option C.