Answer
2.8 y
Work Step by Step
If T is the period of revolution of the earth, r is the orbital radius of the earth, T' is the period of revolution of the asteroid and r' is the radius of orbit of the asteroid, Then
$T^{2}= (1\,year)^{2}=(\frac{4\pi^{2}}{GM})r^{3}$
$(T')^{2}=(\frac{4\pi^{2}}{GM})(r')^{3}$
$\implies \frac{(T')^{2}}{T^{2}}=\frac{(r')^{3}}{r^{3}}=(\frac{r'}{r})^{3}$
But $r'= 2r\implies \frac{r'}{r}=2$
Therefore $(T')^{2}= (\frac{r'}{r})^{3}\times T^{2}=2^{3}\times(year)^{2}$
Or $T'=\sqrt {8\times(year)^{2}}=2\sqrt 2\,y=2.8\,y$
