$\text{Potential energy will be minimum when distance between 2q and 8q is maximum}$

$ \text{Charge q is placed between 2q and 8q}$

$\text{ Let q is placed at a distance x from charge 2q}$

$ U = k( \frac{2q^2}{x} + \frac{8q^2}{9-x} + \frac{16q^2}{9})$

$ \Rightarrow U = kq^2( \frac{2}{x} + \frac{8}{9-x} + \frac{16}{9})$

$\text{For U to be minimum }  \frac{2}{x} + \frac{8}{9-x} \text{should be mimimun}$

$\frac{d}{dx}(\frac{2}{x} + \frac{8}{9-x}) = 0 $

$\Rightarrow x = 3 cm$