From the FBD:
$\Sigma M_C = 0$
$0.25R_{BV} = 0.25(40 \sin 35^{\circ}) + 0.2(40 \cos 35^{\circ})$
$R_{BV} = 49.156 \, \text{kN}$
$\Sigma F_H = 0$
$R_{BH} = 40 \cos 35^{\circ}$
$R_{BH} = 32.766 \, \text{kN}$
$R_B = \sqrt{{R_{BH}}^2 + {R_{BV}}^2}$
$R_B = \sqrt{32.766^2 + 49.156^2}$
$R_B = 59.076 \, \text{kN}$ → shear force of pin at B
$V_B = \tau_B \, A$ → double shear
$59.076 (1000) = \tau_B \bigl\{ \, 2 \left[ \, \frac{1}{4} \pi (20^2) \, \right] \, \bigr\}$
$\tau_B = 94.02 \, \text{ MPa}$ answer