Answer
$1.00000\;\rm cm$
Work Step by Step
We know that the distance the mirror moved is given by
$$\Delta L_2=\dfrac{\Delta m \lambda}{2}$$
Plugging the given;
$$\Delta L_2=\dfrac{(33198) (602.446\times 10^{-9})}{2}=\color{red}{\bf 0.0100000}\;\rm m$$
These zeros are for the appropriate number of significant figures as the author asked.
