First of all: this is not about the physics behind it. It's about the tensor calculation I've written down below. I know this kind of calculation is exhausting but I would be thankful if someone could help me anyway.
In some lecture notes about electro dynamics I read about the electromagnetic tensor and its relation to the magnetic and electric field $\textbf{B}$ respectively $\textbf{E}$. In the following $F_{\mu\nu}$ are the components of the electromagnetic tensor and $\eta$ is the Minkowski metric with signature $(-,+,+,+)$. The components of $\textbf{E}$ are $E_{a}=-F_{0a}$ and the components of $\textbf{B}$ are given by the equality $F_{ab}=\epsilon_{abc}B_{c}$. Then one can compute (this is just for the sake of completeness):
$F_{\mu\nu}F^{\mu\nu}= \\ =F_{\mu\nu}\eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta}\\=F_{0a}\eta^{00}\eta^{ab}F_{0b}+F_{a0}\eta^{ab}\eta^{00}F_{b0}+F_{ab}\eta^{ac}\eta^{bd}F_{cd} \\ =-2F_{0a}\delta^{ab}F_{0b}+F_{ab}\delta^{ac}\delta^{bd}F_{cd}\\=-2\langle\textbf{E},\textbf{E}\rangle+ 2\langle\textbf{B},\textbf{B}\rangle$
where $\langle\,\cdot\,,\,\cdot\,\rangle$ denotes the euclidean dot product. The equality of the first terms is clear to me but I don't see the equality of the second terms.
Edit: My mistake was, that I used the index $c$ twice. I wrote $F_{ab}=\epsilon_{abc}B_{c}$ what is wrong because $c$ is already used as a summation index. Coincidentally the result was the same but nevertheless it was wrong. So here's the right version of it:
$F_{ab}\delta^{ac}\delta^{bd}F_{cd}=\epsilon_{abg}B_{g}\delta^{ac}\delta^{bd}\epsilon_{cdf}B_{f}=\epsilon_{abg}B_{g}\epsilon_{abf}B_{f}=2\delta_{gf}B_{g}B_{f}=2\langle\textbf{B},\textbf{B}\rangle$
Notice that we are in euclidean space $\mathbb{R}^3$ here, which means the position of the indices doesn't matter because you can always use the euclidean metric to "lower" or "upper" indices.
Your problem is $\delta_{bd}\ \delta^{bd}=\delta^b_b=\mathrm{Tr}(\mathrm{Id_3})=3$, not one.