Answer
$0.24\;\rm mm$
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;
$$a=\dfrac{2(600\times 10^{-9})(2.0)}{(2-1)\times 10^{-2}}=\bf 2.4\times 10^{-4}\;\rm m$$
$$a=\color{red}{\bf 0.24}\;\rm mm$$