Assuming the new path of the planet as an elongated ellipse of semi major axis $\frac{r}{2}$ and using kepler's law we get

$\frac{{T_1}^2}{{T_2}^2}=\frac{r^3}{{\displaystyle {\left({\displaystyle \raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^3}}$

$T_2=\frac{T_1}{\sqrt{8}}$

But this is the total time to orbit the sun. To fall down it takes half the total time. So the required time is

$t=\frac{1}{2}\frac{T_1}{\sqrt{8}}=T_1\frac{\sqrt{2}}{8}$