b
As the particles moving in circular orbits, So

\(\frac{{m{v^2}}}{r} = \frac{{16}}{r} + {r^3}\)

Kinetic energy, \(K{E_0} = \frac{1}{2}m{v^2} = \frac{1}{2}\left[ {16 + {r^4}} \right]\)

For fist particle, \(r = 1,\) \({K_1} = \frac{1}{2}\left( {16 + 1} \right)\)

Similarly, for second particle, \(r = 4,\)

\({K_2} = \frac{1}{2}\left( {16 + 256} \right)\)

\(\therefore \,\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{{16 + 1}}{2}}}{{\frac{{16 + 256}}{2}}} = \frac{{17}}{{272}} \simeq 6 \times {10^{ - 2}}\)