Step 1: Write the contraction factor once. With $v = c/\sqrt{2}$, we have $v^2/c^2 = \tfrac{1}{2}$, so the parallel direction shrinks by $\sqrt{1 - v^2/c^2} = \sqrt{1/2} = \dfrac{1}{\sqrt{2}}$.
Step 2: Take the unit stick and keep its two projections separately. The along-motion projection is $\cos 45^\circ = \tfrac{1}{\sqrt{2}}$ and it is multiplied by the shrink factor:
$$L_x = \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} = \frac{1}{2}.$$
Step 3: The transverse projection $\sin 45^\circ = \tfrac{1}{\sqrt{2}}$ is perpendicular to the boost and is untouched, so $L_y = \tfrac{1}{\sqrt{2}}$.
Step 4: Add in quadrature:
$$L = \sqrt{L_x^2 + L_y^2} = \sqrt{\tfrac{1}{4} + \tfrac{1}{2}} = \sqrt{\tfrac{3}{4}} = \frac{\sqrt{3}}{2}.$$
Step 5: The measured rod is shorter than 1 m, as expected for a moving stick.
\[\boxed{L = \frac{\sqrt{3}}{2}\ \text{m}}\]