We can use Kepler's third law to solve this problem. Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun. Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distance from the Sun in kilometers. We can use the following equation to solve for $d$:

$$\frac{T_1^2}{T_2^2} = \frac{d_1^3}{d_2^3}$$

where $T_1$ and $d_1$ are the period of revolution and an average distance of the Earth from the Sun, respectively, and $T_2$ is the period of revolution of the imaginary planet.

Substituting the given values, we get:

$$ \begin{aligned} & \frac{T_{1}}{T_{2}}=\left(\frac{d_{1}}{d_{2}}\right)^{\frac{3}{2}} \\\\ & \Rightarrow\left(\frac{1 \text { year }}{2.83 \text { year }}\right)^{\frac{2}{3}}=\left(\frac{1.5 \times 10^{6} \mathrm{~km}}{d_{2}}\right) \\\\ & \Rightarrow \frac{1}{2}=\frac{1.5 \times 10^{6} \mathrm{~km}}{d_{2}} \\\\ & d_{2}=3 \times 10^{6} \mathrm{~km} \end{aligned} $$