= Solution
A \b[<reflexive Banach space>] is one for which $J_X:X\to X^{**}$ is onto. If $X$ is reflexive, every $\theta\in X^{**}$ has the form $\theta(f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma(X^*,X^{**})$ and $\sigma(X^*,X)$ are precisely the same, proving equality of the topologies.
Conversely, assume those topologies coincide. For any $\theta\in X^{**}$, the <linear functional> $f\mapsto\theta(f)$ on $X^*$ is then weak-star continuous. Its <continuity> at zero gives finitely many $x_1,\ldots,x_m\in X$ and $\epsilon>0$ such that
$$
|f(x_j)|<\epsilon\ \text{for all }j\quad\Longrightarrow\quad|\theta(f)|<1.
$$
If all these evaluations are zero, arbitrary <scalar> multiples of $f$ still satisfy the inequalities, forcing $\theta(f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto(f(x_1),\ldots,f(x_m))$. A <linear functional> on its image extends to $\mathbb R^m$, so for suitable <scalars> $a_j$,
$$
\theta(f)=\sum_{j=1}^ma_jf(x_j)=f\left(\sum_{j=1}^ma_jx_j\right).
$$
Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved
$$
\boxed{X\text{ reflexive}\quad\Longleftrightarrow\quad\sigma(X^*,X)=\sigma(X^*,X^{**}).}
$$
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