None of the standard separation axioms is enough to guarantee $nw(X) \cdot \chi(X)=w(X)$. In fact, in the exercise section of Engelking's chapter on compactness, you can find this construction of a perfectly normal space, known as the Bow-tie Space. Define a topology on the plane by leaving neighborhoods of all points off the x-axis untouched and declaring that a neighborhood of a point $x$ on the x-axis consists of $x$ plus the intersection of the ball of radius $\epsilon$ centered at $x$ with the region bounded by the two lines passing through $x$ and having slopes $\epsilon$ and $-\epsilon$ respectively, for $\epsilon>0$.
The Bow-tie Space is first countable and has a countable network because it's the union of two spaces with a countable base: indeed, the topology induced by this space on both the x-axis and its complement in the plane is the Euclidean one. But its weight is continuum. To see that, let $\mathcal{B}$ be any base. For every $B \in \mathcal{B}$, let $S(B)$ be the set of all points of the x axis such that the vertical line passing through them intersects $B$ in exactly one point. $S(B)$ is a discrete subset of the x-axis (with the Euclidean topology), and hence it's countable. But the $S(B)$'s cover the x axis and hence there have to be continuum many distinct $B$'s.
Ramiro's observations that network weight coincides with weight for compact and metric spaces can be generalized to locally compact and developable spaces, respectively (a "developable space" is a topological space having a sequence $\{\mathcal{U}_n: n < \omega \}$ of open covers such that $\{st(x, \mathcal{U}_n): n < \omega \}$ is a local base at $x$, for every point $x \in X$, where $st(x, \mathcal{U})=\bigcup \{U \in \mathcal{U}: x \in U \}$. The importance of this class lies mainly in a famous topological problem which turned out to be independent of the usual axioms of set theory: does there exist a normal developable non-metrizable space?).
I think it would be interesting to know whether your equality is true for homogeneous spaces.