Given, a toroid with a rectangular
cross-section of inner radius $a$ and outer radius $b$.
Height of the solenoid $=h$
Magnetic field inside a rectangular toroid is given by
$B=\frac{\mu_0 n I}{2 \pi r}$


using the infinitesimal cross-sectional area element,
$d x=h d r$
$\therefore$ Flux passing through the cross-section of toroid.
$\begin{aligned} \phi & =\int B \cdot d x=\int_a^b \frac{\mu_0 n I}{2 \pi r} \cdot(h d r) \\ & \phi=\frac{\mu_0 n I h}{2 \pi} \int_a^b \frac{1}{r} d r \\ \Rightarrow \quad & \phi=\frac{\mu_0 n h I}{2 \pi}\left[\log r l_a^b\right. \\ \phi & =\frac{\mu_0 n h I}{2 \pi}(\log b-\log a) \\ & \phi=\frac{\mu_0 n h I}{2 \pi} \ln \left(\frac{b}{a}\right)\end{aligned}$
Now, self inductance of rectangular toroid,
$L=\frac{n \phi}{I}$
Putting the value of $\phi$, we get
$L=\frac{\mu n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$