$\mathrm{L}_1=100 \mathrm{~cm} \mathrm{\&} \mathrm{T}_1=27^{\circ} \mathrm{C}$
$$
\begin{aligned}
\text { At } 45^{\circ} \mathrm{C}, & \mathrm{L}_2=\mathrm{L}_1+\alpha \mathrm{L}_1 \Delta \mathrm{T} \\
&=100+1.2 \times 10^{-5} \times 100 \times(45-27) \\
&=100.0216 \mathrm{~cm}
\end{aligned}
$$
$\therefore \quad$ Length of $1 \mathrm{~cm}$ mark at $27^{\circ} \mathrm{C}$ on this scale becomes at $45^{\circ} \mathrm{C}=\frac{100.0216}{100} \mathrm{~cm}$.
$\therefore \quad$ Actual length of the steel rod on that day $=\frac{100.0216}{100} \times 63=63.0136 \mathrm{~cm}$
Also, length of the same rod at $27^{\circ} \mathrm{C}$
$$
=63 \times 1=63 \mathrm{~cm}
$$