If the distance of the earth from Sun is 1 . 5 × 10 ^ 6 km , then...

From Kepler's law of periods, we know that

${T}^{2} ∝ {R}^{3} ⇒ {\left(\dfrac{{T}_{1}}{{T}_{2}}\right)}^{2} = {\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{3}$

$⇒ {\left(\dfrac{1}{2 . 83}\right)}^{2} = {\left(\dfrac{1 . 5 \times {10}^{6}}{{R}_{2}}\right)}^{3}$

$⇒ {R}_{2} = \left[{\left(2 . 83\right)}^{2} \times {\left(1 . 5 \times {10}^{6}\right)}^{3}\right]\dfrac{1}{3}$

$= 8\dfrac{1}{3} \times 1 . 5 \times {10}^{6} = 3 \times {10}^{6}km$