Answer
a. It will read low.
b. The percentage error is $0.028\%$.
Work Step by Step
a. At the elevated temperature, the steel tape measure will expand from its calibrated length. The values it shows will be less than the true length, i.e., it will read low.
b. Calculate the percentage error.
$$\frac{\Delta \mathcal{l}}{\mathcal{l}_o}=\alpha \Delta T=(12\times10^{-6}/C)(37C-14C)$$
$$=2.76\times10^{-4}=0.028\%$$
