= 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>,
$$
\boxed{\frac{a_{\rm Mercury}}{a_\oplus}=0.24^{2/3}\simeq0.386,\qquad
\frac{a_{\rm Jupiter}}{a_\oplus}=11.9^{2/3}\simeq5.21.}
$$
Thus the requested orbital-distance estimates are \b[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.
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