$M = 14 ~ \text{kN}\cdot\text{m}$
$I = \dfrac{5b(212^3)}{12} - \dfrac{(5b - 150)(200^3)}{12}$
$I = 3\,970\,053.33b - 3\,333\,333.33b + 100\,000\,000$
$I = 636\,720b + 100\,000\,000$
$f_b = \dfrac{Mc}{I}$
Based on allowable flexural stress of aluminum:
$f_b = 80 / n$
$c = \frac{1}{2}(200) + 6 = 106 ~ \text{mm}$
Thus,
$\dfrac{80}{n} = \dfrac{14(1000^2)(106)}{636\,720b + 100\,000\,000}$
$b = -11.38 ~ \text{mm}$
Based on allowable flexural stress of wood:
$f_b = 10 ~ \text{MPa}$
$c = \frac{1}{2}(100) = 100 ~ \text{mm}$
Thus,
$10 = \dfrac{14(1000^2)(100)}{636\,720b + 100\,000\,000}$
$b = 62.82 ~ \text{mm}$
For stronger section, use $b = 62.82 ~ \text{mm}$ answer
