Answer
$49.4^{\circ}C$
Work Step by Step
We know that $\Delta V = V_0\beta \Delta T$.
For copper, $\beta = 5.1 \times 10^{-5} \ {^{\circ}}C^{-1}$, and $\frac{\Delta V}{V_{0}} = 0.15 \times 10^{-2}.$
Then $\Delta T = \frac{\Delta V/V_0}{\beta} = \frac{0.15 \times 10^{-2}}{5.1 \times 10^{-5}} = 29.4 ^{\circ} C.$
So the final temperature is $(20+29.4) ^{\circ} C = 49.4 ^{\circ}C.$