Since the board is in static equilibrium, there is no rotation around the point where the rope is attached. So the rotation caused by the mass (counter-clockwise) is equal magnitude and in opposite direction of the rotation caused by the weight of the board (clockwise). The weight of the board can be treated like a point for at the center of mass, 0.5 meters in, similar to this picture:
So 0=(moment of board) - (moment of mass)
0= Fboard*(distance to T) - Fmass*(distance to T)
0=(m*10)*0.3 - 3*10*0.2
Solve for m, m=2 kg
Gravity acts on all masses, so at every little point along the bar, gravity is acting, shown by all of the little force arrows. To simplify this for calculations, we can represent all of these little forces as one big force at an object's center of gravity.
Like this: