Slope of grap between $T^2 \;and \;r^3$
$\quad= \large\frac{T^2}{r^3}$
$\quad= \large\frac{\bigg( \Large\frac{2 \pi r}{v_0}\bigg)^2}{r^3}$
$\large\frac{mv_0^2}{r}=\frac{GMm}{r^2}$
$\therefore v_0^2 =\large\frac{GM}{r}$
$\quad = \large\frac{(2 \pi r)^2}{r^3} \times \large\frac{1}{GM}$$r$
$\quad= \large\frac{4 \pi ^2}{GM}$
Hence a is the correct answer.