If the distance of the earth from Sun is $$1.5 \times 10^6$$ km, then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is:

Distance of Earth from Sun = $$1.5 \times 10^6$$ km and the period of the imaginary planet is 2.83 years.

According to Kepler’s Third Law, $$T^2 \propto R^3 \implies \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$$

Substituting Earth’s values $$T_1 = 1$$ year and $$R_1 = 1.5 \times 10^6$$ km into this relation gives $$\left(\frac{2.83}{1}\right)^2 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$

Noting that $$2.83 \approx \sqrt{8}$$ so $$(2.83)^2 \approx 8$$ leads to $$8 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$ and hence $$\frac{R_2}{1.5 \times 10^6} = \sqrt[3]{8} = 2$$

It follows that $$R_2 = 2 \times 1.5 \times 10^6 = 3 \times 10^6 \text{ km}$$

The correct answer is Option C: $$3 \times 10^6$$ km.