Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charge q such that the system is in equilibrium. Then,

$Q=-4q/9, x=d/3$

For Charge Q to be in equilibrium net force on Q is zero.

$\Rightarrow \frac{kqQ}{x^2} = \frac{kQ.4q}{(d-x)^2}$

$\Rightarrow \frac{1}{x^2} = \frac{4}{(d-x)^2}$

$\Rightarrow d-x = 2x$

$\Rightarrow x = \frac{d}{3}$

we have to determine the nature and magnitude of Charge.

Let's take +q charges. The force on it due to +4q charge is directed left wards so force on it due to the third charge should be directed right wards. This implies that the force between +q charge and third charge should be of attractive nature. Thus the nature of the 3rd charge is (−ve).

Taking the third charge to be −Q (say) and then on applying the condition of equilibrium on +q charge 

$\Rightarrow \frac{kqQ}{(d/3)^2} = \frac{kq.4q}{d^2}$

$\Rightarrow 9Q = 4q$

$\Rightarrow Q = \frac{4q}{9}$