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 *f*_{bw} = *f*_{ba} / *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*