In the arrangement depicted in the figure, the system is released...

Let the maximum elongation of the spring be $x$.

At the position of maximum elongation, the velocity of the block is zero. Therefore, the change in kinetic energy of the system is zero ($\Delta K = 0$).

As the block of mass $m$ moves downwards by a distance $x$, the string pulls the top end of the spring upwards by the same distance $x$. Since the bottom end of the spring is fixed to the ground, the spring gets stretched by $x$.

The loss in gravitational potential energy of the block is:
$\Delta U_g = -mgx$

The gain in elastic potential energy of the spring is:
$\Delta U_s = \dfrac{1}{2}kx^2$

According to the principle of conservation of mechanical energy:
$\Delta K + \Delta U_g + \Delta U_s = 0$

$0 - mgx + \dfrac{1}{2}kx^2 = 0$

Since $x \neq 0$ at maximum elongation, we can divide by $x$:
$mg = \dfrac{1}{2}kx$

$x = \dfrac{2mg}{k}$