Solution

In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively:
$$B^{in}_{full} = \mu_0 j x$$
$$B^{out}_{full}=\frac{\mu_0 j h}{2}$$
$$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$
$$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$

Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers:

For $0<x<h/2$:
$$B = \frac{\mu_0 j x}{2}$$

For $x > h/2$:
$$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$