Answer
Please see the work below.
Work Step by Step
We know that
$L=L_{\circ}(1+\alpha \Delta T)$
This can be rearranged as:
$\Delta T=\frac{(\frac{L}{L_{\circ}})-1}{\alpha}$
We plug in the known values to obtain:
$\Delta T=\frac{(\frac{1.000}{0.997})-1}{12\times 10^{-6}}$
$\Delta T=251^{\circ}K$
