From the flexure formula
$f_b = \dfrac{Mc}{I}$
$M = \dfrac{f_b I}{c}$
Without the aluminum plate reinforcement
$I = \dfrac{250(300^3)}{12} = 562\,500\,000 ~ \text{mm}^4$
$c = \dfrac{300}{2} = 150 ~ \text{mm}$
$M_1 = \dfrac{8(562\,500\,000)}{150}$
$M_1 = 30\,000\,000 ~ \text{N}\cdot\text{mm} $
$M_1 = 30 ~ \text{kN}\cdot\text{m}$
With the aluminum plate reinforcement
$I = \dfrac{5(120)(300 + 10 + 10)^3}{12} - \dfrac{[ \, 5(120) - 250 \, ](300^3)}{12} = 850\,900\,000 ~ \text{mm}^4$
$c = \dfrac{300 + 10 + 10}{2} = 160 ~ \text{mm}$
Convert aluminum to wood, use fbw = fba / n
$M_2 = \dfrac{\dfrac{80}{5}(850\,900\,000)}{160}$
$M_2 = 85\,090\,000 ~ \text{N}\cdot\text{mm}$
$M_2 = 85.09 ~ \text{kN}\cdot\text{m}$
Increase in moment capacity
$\Delta M = M_2 - M_1 = 85.09 - 30$
$\Delta M = 55.09 ~ \text{kN}\cdot\text{m}$ answer
