functional analysis – $sigma$-algebra generated by weak topology of Banach space equal Borel $sigma$-algebra

Yes. Let $X$ our Banach space and denote the weak topology $tau’$ and the $sigma$-algebra it generates $sigma(tau’)$. Since a closed ball is closed in the weak topology, all closed balls are in $sigma(tau’)$. Let $B(x,r)$ an open ball. Then
$$B(x,r) = bigcup_n overline{B}(x,r-n^{-1}) in sigma(tau’).$$
Thus all open balls are in $sigma(tau’)$, and since the open balls generate the Borel sigma algebra, $sigma(tau’)$ contains the Borel $sigma$-algebra. The other inclusion is trivial, as the weak topology is weaker than the norm topology.