$$\begin{gathered} 20-\mathrm{mm} \mathrm{diameter} \mathrm{steel} \mathrm{rod} \\ \mathrm{A}=314.16 {\mathrm{mm}}^2 \\ \mathrm{E}=200 \mathrm{GPa} \\ \mathrm{\alpha }=12.0\times {10}^{-6}{/}^{\circ }\mathrm{C} \end{gathered}$$

$$\begin{gathered} {\mathrm{\tau }}_{\mathrm{bolt}}=\frac{{\mathrm{F}}_{\mathrm{bolt}}}{{\mathrm{A}}_{\mathrm{V}}} \\ \therefore {\mathrm{F}}_{\mathrm{bolt}}={\mathrm{\tau }}_{\mathrm{bolt}} {\mathrm{A}}_{\mathrm{V}} \end{gathered}$$

${\mathrm{A}}_{\mathrm{V}}=2{\mathrm{A}}_{\mathrm{bolt}} =2 \left[\frac{\mathrm{\pi }}{4}{(15 \mathrm{mm})}^2\right]=353.43 {\mathrm{mm}}^2$

Since the bolt is in double shear, the area subjected to shear stress is two times the cross-sectional area of the $15-\mathrm{mm}$ diameter bolt.