Two toroids 1 and 2 have total number of turns 200 and 100...

The magnetic field within the turns of toroid is $\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \pi \mathrm{r}}$where, $\mathrm{N}=$ number of turns, $\mathrm{I}=$ current in loops and $\mathrm{r}=$ radius of each turn Here, $\mathrm{N}_1=200, \mathrm{~N}_2=100, \mathrm{r}_1=40 \mathrm{~cm}, \mathrm{r}_2=20 \mathrm{~cm} and current$\mathrm{I}$is same, then$\frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mu_0 \mathrm{~N}_1 $\mathrm{I}$}{2 \pi \mathrm{r}_1} \times \frac{2 \pi \mathrm{r}_2}{\mu_0 \mathrm{~N}_2 $\mathrm{I}$}$Substituting the given values in the above relation, we get$\begin{aligned}\frac{\mathrm{B}_1}{\mathrm{~B}_2}& =\left(\frac{\mathrm{N}_1}{\mathrm{~N}_2}\right)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)\\ & =\left(\frac{200}{100}\right)\left(\frac{20}{40}\right)=2 \times \frac{1}{2}=1 \\ \therefore \quad \mathrm{B}}_{1}: \mathrm{B}}_{2} & =1: 1 \end{aligned}$