A block is kept on the floor of an elevator at rest. The elevator...

The acceleration of the elevator is $a = 5\ \text{m/s}^2$ downwards.

Since the downward acceleration of the elevator is less than the acceleration due to gravity ($a < g$), the block does not lose contact with the floor of the elevator. Thus, the acceleration of the block is the same as that of the elevator, $a = 5\ \text{m/s}^2$.

Using the second equation of motion for the displacement of the block with initial velocity $u = 0$ and time $t = 0.2\ \text{s}$:

$s = ut + \dfrac{1}{2}at^2$

$s = 0 + \dfrac{1}{2} \times 5 \times (0.2)^2$

$s = 2.5 \times 0.04$

$s = 0.1\ \text{m}$

Converting the displacement into centimeters:

$s = 0.1 \times 100\ \text{cm} = 10\ \text{cm}$

Answer: $10\ \text{cm}$