Answer
$14.2\;\rm \mu m$
Work Step by Step
In an interferometer, we know, in condition of constructive interference, that
$$\Delta r=m\lambda$$
where $\Delta r=2(L_2-L_1)$
Hence, in the first case, when you moved M$_2$ away from the spliter,
$$2(L_2-L_1)=m\lambda$$
$$L_2-L_1=\frac{1}{2}m\lambda \tag 1$$
And in the second case, when the other student moved M$_2$ toward the spliter,
$$L_2'-L_1=\frac{1}{2}m'\lambda' \tag 2 $$
Solving (1) for $L_1$l
$$L_1=L_2-\frac{1}{2}m\lambda$$
Plugging into (2);
$$L_2'-\left[L_2-\frac{1}{2}m\lambda\right]=\frac{1}{2}m'\lambda' $$
$$L_2'- L_2+\frac{1}{2}m\lambda =\frac{1}{2}m'\lambda' $$
$$\overbrace{L_2- L_2'}^{\text{M$_2$ net displacement}} =\frac{1}{2}m\lambda-\frac{1}{2}m'\lambda' $$
$$L_2' -L_2=\frac{1}{2}m'\lambda' -\frac{1}{2}m\lambda $$
Plugging the known;
$$ L_2- L_2' =\frac{1}{2}(1200)(632.8)-\frac{1}{2}(1200)(656.5) $$
$$ L_2- L_2'=\bf -14220\;\rm nm$$
$$L_2'-L_2=\color{red}{\bf14.2}\;\rm\mu m$$
