Answer
$d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$
Work Step by Step
We know the following two equations:
$\alpha = \frac{\frac{\Delta L}{L}}{\Delta T}$
And
$L=L_0(1+\alpha \Delta T) $
Using trigonometry to solve for d, it follows:
$d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$
