From the figure:

$V_{max} = \frac{1}{2}P$

$M_{max} = \frac{1}{2}L \, (\frac{1}{2}P) = \frac{1}{4}PL$

**From flexure formula:**

$f_b = \dfrac{6M}{bh^2} = \dfrac{6(\frac{1}{4}PL)}{bh^2}$

$f_b = \dfrac{3PL}{2bh^2}$

$bh = \dfrac{3PL}{2f_b\,h}$

**From shear stress formula:**

$f_b = \dfrac{3V}{2bh} = \dfrac{3(\frac{1}{2}P)}{2\left( \dfrac{3PL}{2f_b\,h} \right)}$

$f_v = \dfrac{f_b \, h}{2L}$ *answer*