Write three equilibrium equations...

$$\begin{gathered} \sum _{} {\mathrm{F}}_{\mathrm{x}}=-{\mathrm{F}}_{\mathrm{AB}}+{\mathrm{C}}_{\mathrm{x}}=0 \sum _{} {\mathrm{F}}_{\mathrm{y}}={\mathrm{C}}_{\mathrm{y}}-50 \mathrm{kN}-15 \mathrm{kN}=0 \\ \sum _{} {\mathrm{M}}_{\mathrm{C}}=\left({\mathrm{F}}_{\mathrm{AB}}\right)(0.4 \mathrm{m})-(50 \mathrm{kN})(0.3 \mathrm{m})-(15 \mathrm{kN})(0.6 \mathrm{m})=0 \end{gathered}$$

and solve for ${\mathrm{F}}_{\mathrm{AB}}$, ${\mathrm{C}}_{\mathrm{x}}$, and ${\mathrm{C}}_{\mathrm{y}}$...

${\mathrm{F}}_{\mathrm{AB}}=60 \mathrm{kN} {\mathrm{C}}_{\mathrm{x}}=60 \mathrm{kN} {\mathrm{C}}_{\mathrm{y}}=65 \mathrm{kN}$

The resultant force in the pin is therefore:

$\left|{\mathrm{F}}_{\mathrm{C}}\right|=\sqrt{{(60 \mathrm{kN})}^2+{(65 \mathrm{kN})}^2} =88.46 \mathrm{kN}$