Answer
$0.5\;\rm mm$
Work Step by Step
We know that the width of the bright central maximum of a single slit is given by:
$$w=\dfrac{2\lambda L}{a}$$
where $a$ is the width of the slit.
So we need to solve for $a$;
$$a=\dfrac{{2\lambda L}}{w}$$
Plugging the known;
$$a=\dfrac{{2\lambda (1.0)}}{4000\lambda}=\bf 5\times 10^{-4} \;\rm m$$
$$a=\color{red}{\bf 0.50}\;\rm mm$$
