As noted by Thierry in his comment, the question isn't posed quite as well as it could be. However, the following example shows that taking duals can result in spaces whose weakly compact subsets can be that are topologically larger than the those of earlier duals. Consider the space $\ell_\infty (S)$ of bounded scalar functions on an infinite set $S$, equipped with the sup supremum norm. First of all, note Note that the Hilbert space $\ell_2 \ell_\infty (S)$ embeds isometrically into has density $\ell_\infty(S)$, but 2^{\vert S\vert}$(with respect to the norm topology), whereas$\ell_2(R)$does not embed \ell_\infty (even isomorphically) into 2^{2^{\vert S\vert }})$ has density $2^{2^{2^{\vert S\vert }}}$; in particular, $\ell_\infty (S)$ if and $\vert R\vert > \vert S\vert$ \ell_\infty (2^{2^{\vert S\vert }})$cannot be isomorphic to one another since every weakly compact subset of$\ell_\infty(S)$has density character not exceeding$\vert S\vert$by a result of 2^{2^{2^{\vert S\vert }}}>2^{\vert S\vert }$. However, H. Rosenthal has shown (for the last claim, see p.230 Theorem 5.1 of Injective Banach spaces and the spaces $L^\infty (\mu)$ for a finite measure $\mu$, Acta Math 124, 1970, 205--248). Observing 205--248) that $\ell_\infty(S)^{\ast\ast}$ is isomorphic to $\ell_\infty (2^{2^{\vert S\vert }})$ (again, see Rosenthal's paper, this time Theorem 5.1 on p.234), we have that })$, hence$\ell_\infty(S)$is not isomorphic to$\ell_\infty(S)^{\ast\ast}$because$\ell_\infty(S)^{\ast\ast}$contains \ell_\infty(S)^{\ast\ast}$. The point of this example is that it is a copy counterexample to the OP's question regardless of whether one considers the Hilbert space canonical embedding of $\ell_2 (2^{2^{\vert S\vert}})$, but X$into$\ell_\infty(S)$does notX^{\ast\ast}$ or one can consider an arbitrary isomorphism between higher duals.
As noted by Thierry in his comment, the question isn't posed quite as well as it could be. However, the following example shows that taking duals can result in spaces whose weakly compact subsets can be topologically larger than the those of earlier duals. Consider the space $\ell_\infty (S)$ of bounded scalar functions on an infinite set $S$, equipped with the sup norm. First of all, note that the Hilbert space $\ell_2 (S)$ embeds isometrically into $\ell_\infty(S)$, but $\ell_2(R)$ does not embed (even isomorphically) into $\ell_\infty (S)$ if $\vert R\vert > \vert S\vert$ since every weakly compact subset of $\ell_\infty(S)$ has density character not exceeding $\vert S\vert$ by a result of H. Rosenthal (for the last claim, see p.230 of Injective Banach spaces and the spaces $L^\infty (\mu)$ for a finite measure $\mu$, Acta Math 124, 1970, 205--248). Observing that $\ell_\infty(S)^{\ast\ast}$ is isomorphic to $\ell_\infty (2^{2^{\vert S\vert }})$ (again, see Rosenthal's paper, this time Theorem 5.1 on p.234), we have that $\ell_\infty(S)$ is not isomorphic to $\ell_\infty(S)^{\ast\ast}$ because $\ell_\infty(S)^{\ast\ast}$ contains a copy of the Hilbert space $\ell_2 (2^{2^{\vert S\vert}})$, but $\ell_\infty(S)$ does not.
It is possible, however, to build an entire zoo of spaces that are isometrically isomorphic to all of their dual spaces - just take $X\oplus_2 X^*$, where $X$ is any reflexive space (here the direct sum is with respect to the $\ell_2$ norm on $\mathbb{R}^2$). Perhaps going off on a bit of a tangent, one can generalise this construction by considering the space $Z = (Y\oplus Y^\ast \oplus Y^{\ast\ast} \oplus Y^{\ast\ast\ast} \oplus \ldots)_{\ell_2}$ for any Banach space $Y$; if $Y$ is reflexive then $Z$ is isometrically isomorphic to all of its duals. If $Y$ is nonreflexive, then for the $m$th and $n$th duals of $Z$, denoted $Z^{m\ast}$ and $Z^{n\ast}$ respectively (here $m$ and $n$ are nonnegative integers, with $Z^{0\ast}=Z$), we still have that $Z^{m\ast}$ and $Z^{n\ast}$ are isometrically isomorphic to subspaces of one another. Moreover, $Z^{m\ast}$ and $Z^{n\ast}$ are isometrically isomorphic to complemented subspaces of one another if neither $m$ or $n$ is $0$ (or if $Y$ is a dual space). However, it still seems to be an open problem (see Plichko and Wojtowicz's paper Note on a Banach space having equal linear dimension with its second dual, Extract Math. 18(3), 2003, p.311--314, for a statement of this problem in the literature) as to whether $X$ and $X^{\ast\ast}$ being isomorphic to complemented subspaces of one another implies that $X$ and $X^{\ast\ast}$ are isomorphic.