Two particles A and B possessing charges q and 2q respectively are...

Let the charge of $C$ be $Q$ and its distance from $A$ be $x$. Since $A$ and $B$ have like charges, they repel each other. For them to remain stationary, $C$ must attract both, meaning $C$ must have a negative charge and be placed between $A$ and $B$.

For the equilibrium of particle $A$, the net force on it must be zero:

$\dfrac{1}{4\pi\epsilon_0} \dfrac{q(2q)}{d^2} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q|Q|}{x^2}$

$\Rightarrow |Q| = \dfrac{2qx^2}{d^2}$

For the equilibrium of particle $B$, the net force on it must be zero:

$\dfrac{1}{4\pi\epsilon_0} \dfrac{q(2q)}{d^2} = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q|Q|}{(d-x)^2}$

$\Rightarrow |Q| = \dfrac{q(d-x)^2}{d^2}$

Equating the two expressions for $|Q|$:

$2x^2 = (d-x)^2$

Taking the square root on both sides:

$\sqrt{2}x = d - x$

$\Rightarrow x(\sqrt{2} + 1) = d$

$\Rightarrow x = \dfrac{d}{\sqrt{2} + 1} = (\sqrt{2} - 1)d$

Substituting the value of $x$ to find $|Q|$:

$|Q| = \dfrac{2q}{d^2} \left( (\sqrt{2} - 1)d \right)^2$

$\Rightarrow |Q| = 2q (2 + 1 - 2\sqrt{2})$

$\Rightarrow |Q| = (6 - 4\sqrt{2})q$

Since the charge must be negative, $Q = -(6 - 4\sqrt{2})q$.