$M = 40 ~ \text{kN}\cdot\text{m}$

$I = \dfrac{15b(270^3)}{12} - \dfrac{(15b - 150)(250^3)}{12}$

$I = 24\,603\,750b - 19\,531\,250b + 195\,312\,500$

$I = 5\,072\,500b + 195\,312\,500$

$f_b = \dfrac{Mc}{I}$

Based on allowable flexural stress of steel:

$f_b = 120 / n$
$c = \frac{1}{2}(250) + 10 = 135 ~ \text{mm}$

Thus,

$\dfrac{120}{15} = \dfrac{40(1000^2)(135)}{5\,072\,500b + 195\,312\,500}$

$b = 94.57 ~ \text{mm}$

Based on allowable flexural stress of wood:

$f_b = 10 ~ \text{MPa}$
$c = \frac{1}{2}(250) = 125 ~ \text{mm}$

Thus,

$10 = \dfrac{40(1000^2)(125)}{5\,072\,500b + 195\,312\,500}$

$b = 60.01 ~ \text{mm}$

For stronger section, use $b = 94.6 ~ \text{mm}$ *answer*