= Solution
If $G$ contains no $K_{3,2}$, every <pair> of <vertices> has at most two <common neighbours>. <Double counting>[Double-counting] a vertex together with an <unordered pair> of its neighbours gives
$$
\sum_v\binom{d(v)}2\leq2\binom n2.
$$
Consequently
$$
\sum_vd(v)^2\leq2n(n-1)+2e.
$$
By Cauchy--Schwarz,
$$
\frac{4e^2}{n}\leq2n(n-1)+2e.
$$
This quadratic inequality implies $e<cn^{3/2}$ for an absolute constant $c$; for example $c=2$ works for every $n\geq1$. Thus
$$
\boxed{\operatorname{ex}(n,K_{3,2})<2n^{3/2}}.
$$
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