 Solution

For planets with negligible mass compared with the Sun, Kepler's third law gives $T^2\propto a^3$, where $a$ is the semi-major axis. Relative to Earth,Thus the requested orbital-distance estimates are about 0.39 and 5.2 times Earth's. Strictly, the period fixes the semi-major axis, not a time-averaged radius of an arbitrary eccentric orbit; identifying the two as a mean distance is the usual elementary orbital-size convention.