The ball tries to expand its volume. But it is kept in the same volume. So it is kept at a constant volume. So the stress that arises is defined as:
$B=\frac{P}{{\displaystyle \frac{∆V}{V}}}$, now we know that change in volume due to thermal expansion, 
$∆V=V\gamma ∆T\Rightarrow \frac{∆V}{V}=\gamma ∆T=3\alpha ∆T$, 
$$\begin{gathered} \Rightarrow B=\frac{P}{{\displaystyle 3\alpha ∆T}} \\ \Rightarrow P=3\alpha ∆T\times B=3\times {10}^{-5}\times 107\times 2\times {10}^{11} \\ \Rightarrow P=6.42\times {10}^8 \mathrm{Pa}\approx 6\times {10}^8 \mathrm{Pa} \end{gathered}$$