According to Kelper's third law (law of periods),

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{T}^2\propto R^3$

where $T$ is time taken by the planet to go once around the sun and $R$ is the semi-major axis (distance) of the elliptical orbit.

$\therefore T^3={k R}^3 \dots$(i)

Where $k$ is a constant of proportionality.

When $R$ becomes $4$ times, let the time period be $T\prime$.

$\therefore {T^{\prime }}^2=k(4R{)}^3 \dots (\mathrm{i}\mathrm{i})$

$\therefore \frac{T^2}{{T^{\prime }}^2}=\frac{1}{64}$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{T}{T^{\prime }}=\frac{1}{8}$

$\mathrm{o}\mathrm{r}\mathrm{ } T^{\prime }=8T$

So, time period becomes $8$ times of previous value.