$R_1 = R_2 = \frac{1}{2}(12)(w)$
$R_1 = R_2 = 6w$
$f_b = 10 \, \text{ksi} = 10,000 \, \text{psi}$
$M = 18w \, \text{lb}\cdot\text{ft}$
$c = 6$
Centroidal moment of inertia of one tube:
$A = \pi r^2 = 0.20$
$r = 0.2523 \, \text{ in}$ hollow portion of the tube was neglected
$\bar I_x = \dfrac{\pi r^4}{4} = \dfrac{\pi (0.2523^4)}{4}$
$\bar I_x = 0.0032 \, \text{in}^4$
Moment of inertia at the center of the section:
$d_1 = 6 \sin 30^\circ = 3 \, \text{in}$
$I_1 = \bar I_x + A{d_1}^2$
$I_1 = 0.0032 + 0.2(3^2)$
$I_1 = 1.8 \, \text{in}^4$
$I_2 = \bar I_x + A{d_2}^2$
$I_2 = 0.0032 + 0.2(6^2)$
$I_2 = 7.2 \, \text{in}^4$
$I = 4I_1 + 2I_2 = 4(1.8) + 2(7.2)$
$I = 21.6 \, \text{in}^4$
$f_b = \dfrac{Mc}{I}$
$10,000 = \dfrac{18w(12)(6)}{21.6}$
$w = 166.7 \, \text{ lb/ft}$ answer
