Answer
$B={\mu_0 \omega Q }/{2\pi R } $
Work Step by Step
Let's assume we can divide the disk into many infinitesimal loops (rings). Each loop contains a charge of $dq$ that is spinning along this loop since the disk is spinning around its center.
We can assume that spinning is a current.
The magnetic field exerted by a current-carrying loop at its center is given by
$$dB_{\rm loop}=\dfrac{\mu_0I}{2r}$$
where $r$ is the radius of the loop and $I=dq_{\rm loop}/dt$,
$$dB_{\rm loop}=\dfrac{\mu_0dq_{\rm loop}}{2rdt}\tag 1$$
The charge area density of the disk is given by
$$\sigma =\dfrac{Q}{A}=\dfrac{Q}{\pi R^2}=\dfrac{dq_{\rm loop}}{A_{\rm loop}}$$
where $R$ is the radius of the disk. So,
$$dq_{\rm loop}=\dfrac{QA_{\rm loop}}{\pi R^2}$$
where $_{\rm loop}= 2\pi r dr$ (we assume that the thickness of one loop is dr), hence
$$dq_{\rm loop}=\dfrac{2\pi rQ dr }{\pi R^2}$$
$$dq_{\rm loop}=\dfrac{2 rQ dr }{ R^2}\tag 2$$
The spinning velocity of the disk is $\omega$ where $\omega=2\pi /t$ and since $\omega$ is constant,
$$dt=\dfrac{2\pi }{\omega}\tag 3$$
Plug (2) and (3) into (1),
$$dB_{\rm loop}=\dfrac{\mu_0 \omega }{4\pi r }\cdot \dfrac{2 rQ dr }{ R^2} $$
$$dB_{\rm loop}=\dfrac{\mu_0 \omega Q dr }{2\pi R^2 } $$
Hence, the net magnetic field of the disk is given by
$$\int_0^B dB_{\rm loop}=\int_0^R \dfrac{\mu_0 \omega Q dr }{2\pi R^2 } $$
$$B=\dfrac{\mu_0 \omega Q }{2\pi R^2 } \int_0^Rdr =\dfrac{\mu_0 \omega Q }{2\pi R^2 } R$$
$$\boxed{B= \dfrac{\mu_0 \omega Q }{2\pi R } }$$
