Let the line have the equation y=mx. It intersects the parabola when x=0 and mx=2x-6x², that is, m=2-6x so x=(2-m)/6.
The curve cuts the x-axis when 2x-6x²=0 that is, when x=0 and ⅓. So if we integrate ydx between these limits we can find the total area under the curve up to the x-axis: we integrate 2x-6x² to get x²-2x³ evaluated between [0,⅓]. This gives us 1/9-2/27=1/27. Half of this is 1/54.
Now consider the area bounded by the line and the curve above the line. We integrate (y-mx)dx for [0,(2-m)/6], which is (2x-6x²-mx)dx. Integrating we get x²-2x³-mx²/2, and putting in the limits we have (2-m)²/36-2(2-m)³/216-m(2-m)²/72=1/54. Let z=2-m, then z²/6-z³/18-(2-z)z²/12=1/9 or 6z²-2z³-3(2-z)z²=4; 6z²-2z³-6z²+3z³=4; z³=4 so z=∛4 and m=2-∛4=0.4126 approx. This is the slope of the line.