As illustrated in the figure, a pulley possesses a radius of 20 cm...

By the principle of conservation of mechanical energy, the loss in gravitational potential energy of the block is equal to the gain in kinetic energy of the block and the pulley, plus the gain in elastic potential energy of the spring.

Let $v$ be the speed of the block and $\omega$ be the angular speed of the pulley after the block has descended by a distance $h$. Since the string does not slip, we have $\omega = \dfrac{v}{R}$.

The energy conservation equation is:
$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}kh^2$

Substituting $\omega = \dfrac{v}{R}$:
$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\left(\dfrac{v}{R}\right)^2 + \dfrac{1}{2}kh^2$

$mgh - \dfrac{1}{2}kh^2 = \dfrac{1}{2}\left(m + \dfrac{I}{R^2}\right)v^2$

Given values:
$m = 1 \text{ kg}$
$g = 10 \text{ m/s}^2$
$h = 10 \text{ cm} = 0.1 \text{ m}$
$k = 50 \text{ N/m}$
$I = 0.2 \text{ kg-m}^2$
$R = 20 \text{ cm} = 0.2 \text{ m}$

Now, substituting the given values into the equation:
$1 \times 10 \times 0.1 - \dfrac{1}{2} \times 50 \times (0.1)^2 = \dfrac{1}{2}\left(1 + \dfrac{0.2}{(0.2)^2}\right)v^2$

$1 - 25 \times 0.01 = \dfrac{1}{2}\left(1 + \dfrac{0.2}{0.04}\right)v^2$

$1 - 0.25 = \dfrac{1}{2}(1 + 5)v^2$

$0.75 = \dfrac{1}{2}(6)v^2$

$0.75 = 3v^2$

$v^2 = \dfrac{0.75}{3} = 0.25$

$v = \sqrt{0.25} = 0.5 \text{ m/s}$