Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can represent the toroid as a long uniform solenoid that was bent to make a circular shape as the toroid.
As we know, a long solenoid has a uniform magnetic field inside it and is parallel to its axis.
And since we will bend it to make a toroid, the magnetic field lines around inside the toroid are circular.
$$\color{blue}{\bf [b]}$$
Let's assume that the current direction is into the page for the inside edge (the small circle) of the toroid and out of the page for the outside edge (the big circle), as shown in the figure below.
Applying Ampere’s law for the closed path;
$$\oint \vec B\cdot d\vec s=\mu_0I_{\rm through}$$
where $I_{\rm through}=NI$
$$\oint \vec B\cdot d\vec s=\mu_0NI$$
As we can see $\vec B$ and $\vec s$ are in the same direction, so
$$B\oint d s=B(2\pi r)=\mu_0NI$$
Thus,
$$\boxed{B=\dfrac{\mu_0NI}{2\pi r}}$$
$$\color{blue}{\bf [c]}$$
It seems, from above, that the toroidal magnetic field is not uniform since $B\propto \dfrac{1}{r}$ where $r$ is the distance from the center of the toroid to radially outward. This means that when $r$ increases $B$ decreases. Hence, the magnetic field is not uniform.