Let us assumed a ring like element of the disc of radius $r$ and thickness $d r$. If $\sigma$ is the surface charge density, then the charge on the element.
$$
d q=\sigma(2 \pi r) d r
$$
Current $d i$ associated with rotating charge $d q$ is
$$
d i=\frac{d q}{T}=\frac{d q \omega}{2 \pi} \quad\left(\because T=\frac{2 \pi}{\omega}\right) \ldots
$$




From Eqs. (i) and (ii), we get
$$
\Rightarrow \quad d i=\sigma \omega r d r
$$
As, the magnetic field $d B$ at centre due to the element,
$$
\begin{aligned}
& d B=\frac{\mu_0 d i}{2 r}=\frac{\mu_0 \sigma \omega r d r}{2 r} \Rightarrow d B=\frac{\mu_0 \sigma \omega}{2} d r \\
& \Rightarrow B_{\text {net }}=\frac{\mu_0 \sigma \omega}{2} \int_0^R d r \quad \Rightarrow B_{\text {net }}=\frac{\mu_0 \sigma \omega R}{2} \\
&
\end{aligned}
$$
Hence, the correct option is (2).