Joint system of RAB after purification of A into R

Let's write the state in terms of its Schmidt decomposition. $$ |psirangle_{AB}=sum_ialpha_i|u_irangle_A|v_irangle_B, $$ ($|u_irangle,|v_irangle$ are orthonormal bases). Then $$ rho_A=sum|alpha_i|^2|u_iranglelangle u_i| $$ and the purification that you would attempt would map $|u_iranglemapsto |u_irangle_A|w_irangle_R$ (state $sum_ialpha_i|u_irangle|w_irangle$). Hence, you can see that the overall state would become $$ sum_ialpha_i|u_irangle_A|v_irangle_B|w_irangle_R, $$ which is GHZ-like, and certainly not separable unless $|psi_{AB}rangle$ was separable.

One way to paraphrase this question is to say that it hypothesizes a state that will not exist in general and then asks what that state is. There is no state of RAB in general that is consistent with both RA and AB being pure; this can only happen when RAB is in a pure product state $|psi_Rrangle|psi_Arangle|psi_Brangle$.

For example, it could be that R, A, and B are all qubits, and we start with AB maximally entangled. This means that A is completely mixed, and purifying A to RA is again maximally entangled. However, there is no state of RAB that is consistent with both RA and AB being maximally entangled, which is a phenomenon commonly known as the monogamy of entanglement.

So what specifically is going wrong in the question? Well, we can certainly consider the state of A in isolation, and consider a purification of this state in RA, but there is no way to do this that is consistent with the existence of B and the state of AB being pure. Purification is not a physical process, for instance, where some quantum channel takes as input the system A and outputs RA that purifies A.