$\Sigma M_A = 0$

$3B_H = 6(20)$

$B_H = 40 ~ \text{kN}$

$\Sigma M_E = 0$

$6B_V + 3(40) = 3(60)$

$B_V = 10 ~ \text{kN}$

Axial Force, *P*_{a}

$P_a = 10 ~ \text{kN compression}$

Moment at *B*

$M_B = 3(20) = 60 ~ \text{kN}\cdot\text{m}$

Maximum compressive stress:

$\sigma_c = \dfrac{P_a}{bd} + \dfrac{6M}{bd^2}$
$\sigma_c = \dfrac{10(1000)}{200(200)} + \dfrac{6(60)(1000^2)}{200(200^2)}$

$\sigma_c = 45.25 ~ \text{MPa}$ *answer*