According to the third Kepler's law,
"\\displaystyle \\frac{a_1^3}{T_1^2} = \\frac{a_2^3}{T_2^2} \\;\\; \\Rightarrow \\; \\Big(\\frac{T_2}{T_1}\\Big)^2 = \\Big(\\frac{a_2}{a_1}\\Big)^3"
where T is a period and a is semi-major axis.
It is given that "a_2=2a_1", so
"\\displaystyle \\Big(\\frac{T_2}{T_1} \\Big)^2 = \\Big(\\frac{2a_1}{a_1} \\Big)^3 = 2^3 = 8"
"\\displaystyle T_2^2 = 8 T_1^2"
"\\displaystyle T_2 = \\sqrt{8} \\,T_1 = 2.828 \\cdot 100 = 282.8 \\; \\textrm{years}"
Answer: 282.8 years
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