Determine the effective acceleration of the ball relative to the lift.

The lift is moving down with an acceleration $a_{lift} = 3 \text{ m/s}^2$.
The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$.
Since both the ball (due to gravity) and the lift are accelerating downwards, the acceleration of the ball relative to the lift is the difference between their accelerations.

The ball is released, meaning its initial velocity relative to the lift is $u = 0 \text{ m/s}$.
The distance the ball needs to fall to hit the floor is $s = 1.7 \text{ m}$.

We use the equation of motion:

Since $u = 0$:

Rearrange the equation to solve for $t$:

Rounding to two significant figures:

The ball will take approximately $\boxed{\text{0.71 s}}$ to hit the floor.