Consider Law of Homogeniety of Dimensions. Now, choose the correct option:

${\mathrm{T}}^2=\frac{4{\pi }^2{\mathrm{r}}^3}{\mathrm{GM}}$

Taking dimensions on both ides, we get

$[\mathrm{T}{]}^2=\frac{[\mathrm{L}{]}^3}{\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\mathrm{M}\right]}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^2\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

$\text{ Now, }{\mathrm{T}}^2=4{\pi }^2{\mathrm{r}}^2$

Taking dimensions on both sides

$[\mathrm{T}{]}^2=[\mathrm{L}{]}^2\therefore \mathrm{LHS}\ne \mathrm{RHS}$

$\text{ Now, }{T}^2=\frac{4{\pi }^2{r}^3}{C}$

Taking dimensions on both sides

$[\mathrm{T}{]}^2=\frac{\left[{\mathrm{L}}^3\right]}{\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^2\right]}\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]$

$\therefore \mathrm{LHS}\ne \mathrm{RHS}$

$\text{ Now, }\mathrm{T}=\frac{4{\pi }^2{\mathrm{r}}^3}{\mathrm{G}}$

$\left[\mathrm{T}\right]=\frac{\left[{\mathrm{L}}^3\right]}{\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]}=\left[{\mathrm{ML}}^0{\mathrm{T}}^2\right]\therefore \mathrm{LHS}$

$\ne \mathrm{RHS}$