The $\mathbf{R}$-vector space $\mathbf{R}^{(\mathbf{R})}$ has dimension $\operatorname{card} \mathbf{R}$ by definition, so it is isomorphic to $\mathbf{R}^{\mathbf{N}}$ (by the Erdős-Kaplansky theorem and because $\operatorname{card}(\mathbf{R}^{\mathbf{N}}) = \operatorname{card} \mathbf{R}$). So $\mathbf{R}^{(\mathbf{R})}$ is isomorphic to the dual of $\mathbf{R}[X]$.
In general, your precise question is equivalent to the following purely set-theoretical question (which seems difficult). By the useful formula, the identity $\operatorname{card} V = \operatorname{dim}_K V$ is equivalent to $\operatorname{card} K \leq \operatorname{dim}_K V$. Let $\kappa = \operatorname{card} K$ and $\lambda = \dim_K V$, and assume $\kappa \leq \lambda$. Does there always exist an infinite cardinal $\alpha$ such that $\kappa^\alpha = \lambda$? (here $\alpha$ is meant to be the dimension of the vector space whose dual is $V$). In general, Stephen's answer shows that there are counterexamples.
EDIT : in order to explain why the question is difficult, consider a field $K$ such that $\operatorname{card} K = \aleph_1$ (for example, one could take $K=\mathbf{Q}((T_i)_{i \in I})$ with $\operatorname{card} I =\aleph_1$). Take a $K$-vector space $V$ of dimension $\aleph_1$. Then $V$ is a dual if and only if there exists $\alpha \geq \aleph_0$ such that $\aleph_1 = (\aleph_1)^\alpha$, which amounts to say that $\aleph_1 = 2^{\aleph_0}$. In other words, $V$ is a dual if and only if the continuum hypothesis holds.