$\Sigma M_A = 0$
$6T = 4(6000) + 12(3000)$
$T = 10,000 ~ \text{lb}$
$\Sigma F_H = 0$
$A_h = T = 10,000 ~ \text{lb}$
$\Sigma F_V = 0$
$A_v = 6000 + 3000 = 9000 ~ \text{lb}$
$P_1 = A_h \cos \theta = 10,000(4/5) = 8000 ~ \text{lb}$
$P_2 = A_v \sin \theta = 9000(3/5) = 5400 ~ \text{lb}$
$P_3 = 6000 \sin \theta = 6000(3/5) = 3600 ~ \text{lb}$
$P_4 = T \cos \theta = 10,000(4/5) = 8000 ~ \text{lb}$
$P_5 = 3000 \sin \theta = 3000(3/5) = 1800 ~ \text{lb}$
$V_1 = A_h \sin \theta = 10,000(3/5) = 6000 ~ \text{lb}$
$V_2 = A_v \sin \theta = 9000(4/5) = 7200 ~ \text{lb}$
$V_3 = 6000 \cos \theta = 6000(4/5) = 4800 ~ \text{lb}$
$V_4 = T \sin \theta = 10,000(3/5) = 6000 ~ \text{lb}$
$V_5 = 3000 \cos \theta = 3000(4/5) = 2400 ~ \text{lb}$
$P_1 + P_2 = 8000 + 5400 = 13,400 ~ \text{lb}$
$V_2 - V_1 = 7200 - 6000 = 1200 ~ \text{lb}$
At Point B
$\sigma_a = \dfrac{P}{A} = \dfrac{13,400}{6(10)}$
$\sigma_a = 223.33 ~ \text{psi compression}$
$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(6000)(12)}{6(10^2)}$
$\sigma_f = 720 ~ \text{psi at extreme tension and compression fibers}$
$\sigma_B = \sigma_a + \sigma_f = 223.33 + 720$
$\sigma_B = 943.33 ~ \text{psi}$
At Point C
$\sigma_a = \dfrac{P}{A} = \dfrac{9800}{6(10)}$
$\sigma_a = 163.33 ~ \text{psi compression}$
$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(12,000)(12)}{6(10^2)}$
$\sigma_f = 1440 ~ \text{psi at extreme tension and compression fibers}$
$\sigma_C = \sigma_a + \sigma_f = 163.33 + 1440$
$\sigma_C = 1603.33 ~ \text{psi}$
Thus, maximum compressive stress is that at C
$\sigma_{max} = 1603.33 ~ \text{psi}$ answer
