Maximum moment will occur at the midspan
$M_{max} = \dfrac{w_oL^2}{8} = \dfrac{300(20^2)}{8}$
$M_{max} = 15\,000 ~ \text{lb}\cdot\text{ft}$
From Solution 105:
$\bar{y} = 7.1 ~ \text{in}$
$I_{NA} = \frac{3487}{3} ~ \text{in}^4$
$f_b = \dfrac{Mc}{I}$
At the extreme wood fiber ($c = \bar{y}$)
$f_{bw} = \dfrac{M_{max} \, \bar{y}}{I_{NA}}$
$f_{bw} = \dfrac{15\,000(7.1)(12)}{\frac{3487}{3}}$
$f_{bw} = 1099.51 ~ \text{psi}$ answer
At the extreme steel fiber ($c = 10.5 - \bar{y}$)
$\dfrac{f_{bs}}{n} = \dfrac{M_{max}(10.5 - \bar{y})}{I_{NA}}$
$\dfrac{f_{bs}}{20} = \dfrac{15\,000(10.5 - 7.1)(12)}{\frac{3487}{3}}$
$f_{bs} = 10\,530.54 ~ \text{psi}$ answer
