To satisfy the $50 \mathrm{MPa}$ allowable stress limit, the pin must provide a shear area of at least $1,769.2 {\mathrm{mm}}^2$. From the side view of the support at $\mathrm{C}$, the pin at $\mathrm{C}$ is observed to be in double shear, meaning that shear stresses will be developed on two surfaces.

Therefore, the shear area for the pin is twice the cross-sectional area:

$$\begin{gathered} {\mathrm{A}}_{\mathrm{V}}={\displaystyle 2 {\mathrm{A}}_{\mathrm{pin}}=2(\mathrm{\pi }/4)({\mathrm{d}}_{\mathrm{pin}}{)}^2}=1,769.2 {\mathrm{mm}}^2 \\ \therefore {\mathrm{d}}_{\mathrm{pin}}=\mathbf{33}\mathbf{.6 }\mathbf{mm} \end{gathered}$$