1. Use Kepler's third law
   For planets revolving around the same Sun, $T^2 \propto r^3$ So, $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$

2. Substitute the given values
   For Earth: $T_1 = 1\text{ year}, \quad r_1 = 1.5\times 10^8\text{ km}$ (Here, the standard Earth-Sun distance must be $1.5\times 10^8$ km; $1.5\times 10^6$ km is evidently a typo, otherwise no option is sensible.)

For the imaginary planet: $T_2 = 2.83\text{ years}$

Thus, $\left(\frac{r_2}{r_1}\right)^3 = (2.83)^2$ $\frac{r_2}{r_1} = (2.83)^{2/3}$

3. Evaluate the power
   Since, $2.83 \approx \sqrt{8} = 2.828$ Therefore, $(2.83)^2 \approx 8$ So, $\left(\frac{r_2}{r_1}\right)^3 \approx 8$ $\frac{r_2}{r_1} \approx 2$

Hence, $r_2 = 2r_1 = 2\times 1.5\times 10^8$ $r_2 = 3\times 10^8\text{ km}$

4. Compare with options
   None of the given options match $3\times 10^8$ km exactly. However, if the intended Earth-Sun distance were written as $1.5\times 10^7$ km, then $r_2 = 3\times 10^7\text{ km}$ which matches Option B.

So the problem statement/options contain an inconsistency, but the physically correct scaling gives the corresponding choice as:

$\boxed{3\times 10^7\text{ km}}$ if one follows the option pattern.

Therefore, the best answer among the options is B.