Magnetic flux linked between two coils placed co-axially

G. Smith gave the answer (brute force method) but there is a trick to do the problem other way around which is usually done on UG level. Pass current in bigger coil and find the mutual inductance ($M$) between the two coil. The flux linked due to current in smaller coil will simply the product of $M$ and current in smaller coil. I would give one simple calculation of $M$ $$phi_{mathrm{link}}=intvec{B}•dvec{a}$$ $$approxvec{B}•int dvec{a}$$ Magnetic field stays constant over the whole area of smaller ring since it's area is very small $$frac{mu_{0}}{2}frac{I R^2}{(z^2+R^2)^{3/2}}pi r^2$$This is equal to $M$ times current in larger coil you can now proceed further.

Refinement: if you want to decrease the distance between the two coils so our area approximation doesn't hold anymore then you can simply go for multiple expansion. And if you're champion of integration Neumann formuale is there for you $$M=frac{mu_0}{4pi}oint frac{dvec{l}_1•dvec{l}_2}{mathcal{r}}$$