Flexural stress developed:
$M = \dfrac{EI}{\rho}$

$f_b = \dfrac{Mc}{I} = \dfrac{(EI/\rho)c}{I}$

$f_b = \dfrac{Ec}{\rho} = \dfrac{200000(0.80/2)}{300}$

$f_b = 266.67 \, \text{MPa}$

Minimum diameter of pulley:

$f_b = \dfrac{Ec}{\rho}$
$400 = \dfrac{200000(0.80/2)}{\rho}$

$\rho = 200 \, \text{mm}$

Diameter, d = 400 mm *answer*