Answer
$$A_{BE} = 285.3 \space mm^2$$
Work Step by Step
1) Reactions and internal forces
$\bullet$Start by examining the link CD, applying equilibrium, we get:
$\sum{M_D} = 0; \space\space 4*0.15 + 4*0.25 - V_C * 0.25 \space \Rightarrow \space
V_C = 6.4 \space kN \uparrow$
$\bullet$ Now examine the portion of the structure to the left of the internal hinge at C
Applying the equilibrium of moments around point A we can find force in link BE
$\sum{M_A} = 0; \space\space 4*0.35 +6.4 * 0.45 - BE * 0.15 = 0$
$BE = 28.53 \space kN$
2) Average Normal Stress:
Normal Stress equation:
$\sigma_{AVG} = \frac{N_{INT}}{A} \space\Rightarrow\space A = \frac{N_{INT}}{\sigma_{AVG}}$
$A = \frac{28.53 * 10^3 \space N}{100 * 10^6 \space Pa} = 2.853 *10^{-4}\space m^2 = \boxed{285.3 \space mm^2} \space\space\space\leftarrow ANS$
