$\newcommand{\Z}{\mathbb{Z}}$ Take the fibration $K(\Z,2) \hookrightarrow * \to K(\Z,3)$. Then $d_3^{0,2}$ is an isomorphism since this is the only way to get rid of $H^2(K(\Z,2))$ and to kill $H^3(K(\Z,3))$. Therefore $i \in \Z[i]=H^*(K(\Z,2))$ is sent under $d^3$ to a generator which has to be a fundamental class $j$ of $H^3(K(\Z,3))$. Therefore $d_3^{0,4}$ is the multiplication by $2$ map: $d_3^{0,4}(i^2)=j \otimes i +(-1)^{2+2} i \otimes j \mapsto (-1)^{2*3} ij +i j=2 ij \in \Z\langle ij \rangle=H^3(K(\Z,3),\Z)=E_3^{3,2}$ . Here $\mapsto$ is the cup product map and $\langle \rangle$ denotes 'module generated by'.
$d_3^{3,2}$ is the zero map since $d_3(j \otimes i)=-j\otimes j \mapsto j^2=0$ since $j$ is of odd degree. Therefore $E_4^{3,2}=\Z/2$ and there is nothing to kill or get rid of it. Therefore $H^5(*) \neq 0$. Contradiction.
What is the error in calculating this spectra sequence.
This calculation was done while doing the spurious calculation in Another way to compute $\pi_4(S_3)$: contradiction in spectral sequence calculation
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The error is that $j^2\neq 0$. The fact that $j$ has odd degree tells you that $j^2=-j^2$, but this just means $2j^2=0$, not $j^2=0$. In fact, your computation gives a proof that $j^2$ cannot be $0$ and hence is an element of order $2$ in $H^6(K(\mathbb{Z},3);\mathbb{Z})$.