Answer
$0.12\;\rm mm$
Work Step by Step
We have here a single-slit board that we need to find its width.
We knowو in the single slit experiment, that the position of the dark fringe is given by
$$y_p=\dfrac{p\lambda L}{a}$$
So the distance between the first and the third dark fringe is given by
$$\Delta y=y_3-y_1=\dfrac{3\lambda L}{a}-\dfrac{ \lambda L}{a}$$
$$\Delta y =\dfrac{2\lambda L}{a} $$
Solving for $a$;
$$a =\dfrac{2\lambda L}{\Delta y} $$
Plugging the known;
$$a =\dfrac{2(589\times 10^{-9})(0.75)}{(7.5\times 10^{-3})}=\bf 1.178\times 10^{-4}\;\rm m $$
$$a\approx\color{red}{\bf 0.12}\;\rm mm$$
