To find the required pushing force, we analyze the forces acting on the box. Since the box moves at a constant speed, its acceleration is zero, and the net force on it is zero according to Newton's First Law.

We define a coordinate system with the x-axis parallel to the incline (positive upwards) and the y-axis perpendicular to it.

The forces acting on the box are:

1. The pushing force, $F_p$, directed up the incline.
2. The gravitational force, $F_g = mg$, acting vertically downwards.
3. The normal force, $F_N$, perpendicular to the incline.
4. The kinetic friction force, $F_f$, opposing motion, directed down the incline.

We resolve the gravitational force into components parallel ($mg\sin\theta$) and perpendicular ($mg\cos\theta$) to the incline. The angle of inclination $\theta$ is determined by the plank's length $L$ and height $h$.

Applying Newton's First Law ($\Sigma \vec{F} = 0$): Sum of forces in the y-direction (perpendicular to incline):

$$\Sigma F_y = F_N - mg\cos\theta = 0$$ $$F_N = mg\cos\theta$$

Sum of forces in the x-direction (parallel to incline):

$$\Sigma F_x = F_p - mg\sin\theta - F_f = 0$$

The kinetic friction force is $F_f = \mu_k F_N = \mu_k mg\cos\theta$. Substituting this into the x-direction equation:

$$F_p - mg\sin\theta - \mu_k mg\cos\theta = 0$$

Solving for the pushing force $F_p$:

$$F_p = mg\sin\theta + \mu_k mg\cos\theta$$ $$F_p = mg(\sin\theta + \mu_k \cos\theta)$$

From the problem geometry, with plank length $L=3.0$ m and height $h=1.3$ m:

$$\sin\theta = \frac{h}{L}$$ $$\cos\theta = \frac{\sqrt{L^2 - h^2}}{L}$$

Substituting these into the expression for $F_p$:

$$F_p = mg \left( \frac{h}{L} + \mu_k \frac{\sqrt{L^2 - h^2}}{L} \right) = \frac{mg}{L} \left( h + \mu_k \sqrt{L^2 - h^2} \right)$$

Now, we substitute the given values: $m = 200$ kg, $g = 9.8$ m/s², $L = 3.0$ m, $h = 1.3$ m, and $\mu_k = 0.3$.

$$F_p = \frac{(200 \text{ kg})(9.8 \text{ m/s}^2)}{3.0 \text{ m}} \left( 1.3 \text{ m} + 0.3 \sqrt{(3.0 \text{ m})^2 - (1.3 \text{ m})^2} \right)$$ $$F_p = \frac{1960}{3.0} \left( 1.3 + 0.3 \sqrt{9.0 - 1.69} \right) \text{ N}$$ $$F_p = \frac{1960}{3.0} \left( 1.3 + 0.3 \sqrt{7.31} \right) \text{ N}$$ $$F_p \approx 653.33 (1.3 + 0.3 \times 2.704) \text{ N}$$ $$F_p \approx 653.33 (1.3 + 0.8112) \text{ N}$$ $$F_p \approx 653.33 (2.1112) \text{ N}$$ $$F_p \approx 1380 \text{ N}$$