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First, that's not a rotational problem.  It's a weight balance or about lever arms.

For simple problems like this, it's good to draw a picture:

<img src=https://physics.codidact.com/uploads/nypvhjri8nssuq75j4lo419tc57x>

Since the board is 6 m long and 2 m are sticking out on one side (right in this drawing), 4 m must be on the other side (left) of the fulcrum.  The left side is &frac23; the mass of the whole board, and the right side the other &frac13;.  That results in the masses shown.

For moment arm purposes, we can consider each part of the board to be a point mass at its center, shown by the two dots.  The distances from the fulcrum are half the length of each sections, so 2 m at left and 1 m at right.

The left moment arm is holding the board onto the building with (36.7 kg)(2 m)g, while the right end is trying to flip the board with (18.3 kg)(1 m)g.  That means with nothing else on the board, there is a moment of (73.3 kg)g - (18.3 kg)g = (55 kg)g holding the board to the building.

The question now is how far to the right can you add 40 kg to exactly balance the board?  If x is the distance to the right of the fulcrum to apply the 40 kg mass, then the moment of that mass will be (40 kg)(x m)g.  It should be obvious that x = (55 kg)g / (40 kg)g = 1.375.

So the answer is, the board will start to tip when the 40 kg worker gets 1.375 m out from the edge.

You may be wondering why I multiplied expressions by g, the acceleration due to gravity.  First, a moment arm is force times distance, not mass times distance.  All too often people get sloppy and assume a 1 g environment like here on the surface of the earth.  Second, I wanted to show that g cancels out.  Perhaps that's intuitive, but it means that the answer is the same on Mars or the moon, where g is different.  If the masses were compressing springs, for example, g would matter and can't be cancelled out.  That's essentially the difference between a scale and a balance.