The self-inductance L of a toroid having circular cross-section of radius $r$ and major radius $R$ is ($r \ll R$)

The magnetic field inside a toroid of major radius $R$ is given by $B = \dfrac{\mu_0 N I}{2 \pi R}$, assuming $r \ll R$ so that the field is approximately uniform across the cross-section.

The cross-sectional area of the toroid is $A = \pi r^2$.

The magnetic flux through a single turn is $\phi = B A = \left(\dfrac{\mu_0 N I}{2 \pi R}\right) (\pi r^2) = \dfrac{\mu_0 N I r^2}{2 R}$.

The total flux linkage for $N$ turns is $\Phi = N \phi = \dfrac{\mu_0 N^2 I r^2}{2 R}$.

The self-inductance is $L = \dfrac{\Phi}{I} = \dfrac{\mu_0 N^2 r^2}{2 R}$.

Answer: $\dfrac{\mu_0 N^2 r^2}{2R}$