As we know by Kepler's law
Square of time period is always proportional to the cube of time period
it is given by
![(T_1^2)/(T_2^2) = (r_1^3)/(r_2^3)](https://img.qammunity.org/2019/formulas/physics/college/6lm6wub1a0t1kywe5mfk1h6r2or43ynezv.png)
here we have
T1 = 3000 years
r1 = distance of comet
T2 = 1 year
r2 = 1AU
now by above equation we will have
![(3000^2)/(1^2) = (r_1^3)/(1^3)](https://img.qammunity.org/2019/formulas/physics/college/m44arqbncq7nt2mf5e5uh3gq4ygwvb0p3l.png)
![9*10^6 = r_1^3](https://img.qammunity.org/2019/formulas/physics/college/vy7df1iu9mxs0yf91gn2zwimmqay358ovt.png)
![r_1 = 208 AU](https://img.qammunity.org/2019/formulas/physics/college/2dcsvugjw9af20iyhrcopk8lqaf6meir5x.png)
so the distance of comet will be 208 AU.