Determine the acceleration of the box down the smooth inclined plane.

The component of gravitational force acting parallel to the incline causes the acceleration.

$$a = g \sin\theta$$

Given $g = 10\, \text{m/s}^2$ and $\theta = 30^\circ$:

$$a = 10\, \text{m/s}^2 \times \sin(30^\circ)$$ $$a = 10\, \text{m/s}^2 \times 0.5$$ $$a = 5\, \text{m/s}^2$$

The box starts from rest, so initial velocity $u = 0\, \text{m/s}$. The distance slid is $s = 6\, \text{m}$.
Using the equation $v^2 = u^2 + 2as$:

$$v^2 = (0\, \text{m/s})^2 + 2(5\, \text{m/s}^2)(6\, \text{m})$$ $$v^2 = 0 + 60\, \text{m}^2/\text{s}^2$$ $$v = \sqrt{60}\, \text{m/s}$$ $$v \approx 7.74596\, \text{m/s}$$

Correct to one decimal place:

$$\boxed{v = 7.7\, \text{m/s}}$$

Momentum $p = mv$.
Given mass $m = 20\, \text{kg}$ and velocity $v = \sqrt{60}\, \text{m/s}$:

$$p = 20\, \text{kg} \times \sqrt{60}\, \text{m/s}$$ $$p \approx 20\, \text{kg} \times 7.74596\, \text{m/s}$$ $$p \approx 154.9192\, \text{kg m/s}$$

Correct to one decimal place:

$$\boxed{p = 154.9\, \text{kg m/s}}$$