Solution

a) As in the previous problem, we will use the property of the vector product and the formula for the field inside a cylindrical wire. Consider a point at distances $r_1$ and $r_2$ from the centers of the cylinders:
$$\hat{B} = \frac{\mu_0 [\hat{r}_1 \times \hat{j}]}{2} - \frac{\mu_0 [\hat{r}_2 \times \hat{j}]}{2} = \frac{\mu_0 [\hat{a} \times \hat{j}]}{2}$$
$$B = \frac{\mu_0 a j}{2}$$
b) The detailed solution to this point repeats in all its main points the solution to problem 6.2.14, which I recommend you read. However, it should be mentioned that here it is more convenient to express the answer in terms of a sine, since the induction lines inside are perpendicular to the vector $\hat{a}$.
$$i = j a sin \varphi$$
$$j = \frac{2B_0}{\mu_0 a}$$
$$i = \frac{2B_0}{\mu_0}sin\varphi$$
$$i_{max} = \frac{2B_0}{\mu_0}$$