A block rests on the floor of a stationary elevator. The elevator...

The downward acceleration of the elevator is $a = 12 \text{ m/s}^2$.

Since the downward acceleration of the elevator is greater than the acceleration due to gravity ($a > g$), the normal force becomes zero and the block loses contact with the floor of the elevator.

The block will undergo free fall under gravity with an acceleration $g = 10 \text{ m/s}^2$.

Using the second equation of motion for the block:

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

Given the initial velocity $u = 0$ and time $t = 0.2 \text{ s}$:

$s = 0 + \dfrac{1}{2} \times 10 \times (0.2)^2$

$s = 5 \times 0.04 = 0.2 \text{ m}$

$s = 20 \text{ cm}$