a) The free body diagram is
To check for equilibrium, I have to show that RC acts at the centroid of the load. But this is obvious, since the centroid of a triangular area is one-third of the height, which is L.
b) To find the position of the maximum normal stress, I need the maximum bending moment. To get the position of the maximum moment, I must write the moment equation, take its derivative (which is the shear), and equate that to zero. Then I plug that value into the moment equation to compute the bending moment. That can then be used to find the stress at that position.
The equation for the load is 
.
M and V are both zero at the ends, as can be seen by mentally drawing free body diagrams. Using this and
This equation is valid from the left end
up to point C. Substituting x = 2L/3, I find that the shear is
just to the left of C. At C the reaction drives the shear up, yielding
a value of 
.
The shear passes through a discontinuity at x = 2L/3. The change in the bending moment is the area under the shear curve. Thus, the bending moment reaches a minimum (maximum absolute value) at x = 2L/3; then the positive shear area begins to have an effect, driving the bending moment to zero at x = L. The value of the bending moment at x = 2L/3 is found by deriving the moment equation and substituting this value of x:
This is valid from x = 0 to a point just left of point C. But the moment has to be continuous since there is no external moment, and we know that the moment at the right end is zero...
The moment of inertia of the cross section
is 
and c
= a/2:
Q.E.D.