Kepler's third law states that the square of the period of...

Here gravitational force is equal to centripetal force required for motion of planet.
Hence,
$\dfrac{GMm}{{r}^{2}} = \dfrac{m{v}^{2}}{r}$
${v}^{2} = \dfrac{GM}{r}$
Time period is given by,
$T = \dfrac{2\pir}{v}$
${T}^{2} = \dfrac{4{\pi}^{2}{r}^{2}}{{v}^{2}}$
${T}^{2} = \dfrac{4{\pi}^{2}{r}^{2}}{\left(\dfrac{Gm}{r}\right)} = \dfrac{4{\pi}^{2}{r}^{3}}{GM}$

We have given,
${T}^{2} = k{r}^{3}$
On comparing,
$k = \dfrac{4{\pi}^{2}}{GM} ⇒ GMk = 4{\pi}^{2}$