Answer
${\bf 2.0}\;\rm cm$
Work Step by Step
We know that the magnetic field of a coil of $N$ turns at its center is given by
$$B_{\rm center}=\dfrac{\mu_0N I}{2 R}$$
where $N=L/2\pi R$, so
$$B_{\rm center}=\dfrac{\mu_0L I}{4\pi R^2}$$
Hence,
$$R^2 =\dfrac{\mu_0L I}{4\pi B_{\rm center}}$$
where the diameter is $D=2R$, so $R=D/2$,
$$D^2 =\dfrac{4\mu_0L I}{4\pi B_{\rm center}}$$
$$D =\sqrt{\dfrac{ \mu_0L I}{ \pi B_{\rm center}}}$$
Plug the known;
$$D =\sqrt{\dfrac{ (4\pi\times 10^{-7})(1)(1)}{ \pi (1\times 10^{-3})}}$$
$$D=\color{red}{\bf 2.0}\;\rm cm$$
