(c) Here, $\mathrm{\Delta}l = 80.3 - 80.0 = 0.3\ $cm
$l = 80\ cm,\ \alpha = 12\ \times 10^{- 6}℃^{- 1}$
Rise in temperature $\mathrm{\Delta}T = \frac{\mathrm{\Delta}l}{l\alpha}$
$\mathrm{\Delta}T = \ \frac{0.3}{80 \times 12 \times 10^{- 6}} = 312.5℃$