Answer
$500\;\rm nm$
Work Step by Step
We know, in a double-slit experiment, that the position of the dark fringes is given by
$$y'_m=\left(m+\frac{1}{2}\right)\dfrac{\lambda L}{d}$$
where $L$ is the separation distance between the slit and the screen, $d$ is the distance between the two slits, and $m$ is the number of the dark fringe up or down the central maxima.
Thus, the separation distance between the first dark and the fifth dark is given y
$$y'_5-y'_1=\left(5+\frac{1}{2}\right)\dfrac{\lambda L}{d}-\left(1+\frac{1}{2}\right)\dfrac{\lambda L}{d}$$
$$y'_5-y'_1=(5.5-1.5)\dfrac{\lambda L}{d} $$
Solving for $\lambda$;
$$\lambda=\dfrac{(y'_5-y'_1)d}{4L}$$
Plugging the known;
$$\lambda=\dfrac{(6\times 10^{-3})(0.20\times 10^{-3})}{4(0.60)}=500\times 10^{-9}\;\rm m$$
$$\lambda=\color{red}{\bf 500}\;\rm nm$$
