a b c d

a

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    $\Rightarrow \frac{kqQ}{(d/3)^2} = \frac{kq.4q}{d^2}$   $\Rightarrow 9Q = 4q$   $\Rightarrow Q = \frac{4q}{9}$