A toroid has a non ferromagnetic core of inner radius 24   cm and outer radius 25   cm , around…

Here, $r_1\mathrm{ }=\mathrm{ }24 \mathrm{cm}=24\times {10}^{-2}\mathrm{ }\mathrm{m}$

$r_2\mathrm{ }=\mathrm{ }25 \mathrm{cm}=25\times {10}^{-2}\mathrm{ }\mathrm{m}$
$N=4900$ turns, $I=12\mathrm{ }\mathrm{A}$

Now, $r=\frac{r_1+r_2}{2}=24.5\times {10}^{-2}\mathrm{ }\mathrm{m}$
Therefore, magnetic field will be

$$\begin{gathered} B=\frac{{\mu }_{0}NI}{2\mathrm{\pi r}} \\ =\frac{\left(4\pi \times {10}^{-7}\right)\left(4900\right)\left(12\right)}{2\pi \times \left(24.5\times {10}^{-2}\right)} \\ =4800\times {10}^{-5} \\ \approx 48 \mathrm{mT} \end{gathered}$$