Answer
$0.2894\;\rm mm$
Work Step by Step
We have here two wavlengths $\lambda_1=589.6$ nm and $\lambda_2=589.0$ nm where $\lambda_2\lt \lambda_1$.
We need to find the distance mirror 2 would be moved so that one wavelength has produced one more new maximum than
the other wavelength.
So,
$$\Delta L_2=\dfrac{m\lambda_1}{2}=\dfrac{(m+1)\lambda_2}{2}$$
Thus,
$$m\lambda_1=m\lambda_2+\lambda_2$$
$$m(\lambda_1- \lambda_2)=\lambda_2$$
Hence,
$$m=\dfrac{\lambda_2}{\lambda_1- \lambda_2}\tag 1$$
So the distance the mirror moved is given by
$$\Delta L_2=\dfrac{ m \lambda_1}{2}$$
Plugging from (1);
$$\Delta L_2=\dfrac{\lambda_2}{\lambda_1- \lambda_2}\dfrac{ \lambda_1}{2}$$
Plugging the known;
$$\Delta L_2=\dfrac{589.0}{589.6- 589.0}\dfrac{ 589.6}{2}=\bf 289395\;\rm nm$$
$$\Delta L_2= \color{red}{\bf 2.894\times 10^4}\;\rm m$$