This is a question related to the Einstein summation convention. I have made an attempt at answering this question and have been incorrect. I'd really like for someone to help me understand why it is that my method fails. The following is my attempt (it should be noted here that $\delta\_{ij}$ is the Kronecker-delta).

We may fix certain variables, and expand over others. For this purpose, we will start by fixing $j$, $k$, $m$ and $n$, and letting $i$ range from $1$ to $3$. We obtain
\begin{equation}
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n=\delta\_{1j}\delta\_{jk}\delta\_{k1}\delta\_{mn}a\_mb\_n+\delta\_{2j}\delta\_{jk}\delta\_{k2}\delta\_{mn}a\_mb\_n+\delta\_{3j}\delta\_{jk}\delta\_{k3}\delta\_{mn}a\_mb\_n
\end{equation}
Now we will let $j$ range from $1$ to $3$. We note therefore that $j$ will range from $1$ to $3$ for each term of $(4)$. Therefore, if we omit terms which equate to $0$, we obtain
\begin{equation\*}
\begin{split}
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n & =\delta\_{11}\delta\_{1k}\delta\_{k1}\delta\_{mn}a\_mb\_n+\delta\_{22}\delta\_{2k}\delta\_{k2}\delta\_{mn}a\_mb\_n+\delta\_{33}\delta\_{3k}\delta\_{k3}\delta\_{mn}a\_mb\_n \\
& =\delta\_{1k}\delta\_{k1}\delta\_{mn}a\_mb\_n+\delta\_{2k}\delta\_{k2}\delta\_{mn}a\_mb\_n+\delta\_{3k}\delta\_{k3}\delta\_{mn}a\_mb\_n
\end{split}
\end{equation\*}
We now let $k$ range from $1$ to $3$, still omitting those terms which equate to $0$, and we notice that
\begin{equation\*}
\begin{split}
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n & =\delta\_{11}\delta\_{11}\delta\_{mn}a\_mb\_n+\delta\_{22}\delta\_{22}\delta\_{mn}a\_mb\_n+\delta\_{33}\delta\_{33}\delta\_{mn}a\_mb\_n \\
& =\delta\_{mn}a\_mb\_n+\delta\_{mn}a\_mb\_n+\delta\_{mn}a\_mb\_n
\end{split}
\end{equation\*}
Next, we let $m$ range from $1$ to $3$. Since all other indices have been eliminated at this point, we do not need to be concerned with them. We have that
$$
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n=\delta\_{1n}a\_1b\_n+\delta\_{2n}a\_2b\_n+\delta\_{3n}a\_2b\_n
$$
And finally we let $n$ range from $1$ to $3$. We omit terms which equate to $0$, and hence we obtain
\begin{equation\*}
\begin{split}
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n & =\delta\_{11}a\_1b\_1+\delta\_{22}a\_2b\_2+\delta\_{33}a\_2b\_3 \\
& =a\_1b\_1+a\_2b\_2+a\_3b\_3 \\
\end{split}
\end{equation\*}
Hence we may introduce the dummy variable $r$, and, noting that $a\_1b\_1+a\_2b\_2+a\_3b\_3=a\_rb\_r=\vec{a}\cdot\vec{b}$, we have that
$$
\delta\_{ij}\delta\_{jk}\delta\_{ki}\delta\_{mn}a\_mb\_n=\vec{a}\cdot\vec{b}
$$

This is the answer I arrived at, although I am aware that the answer is supposed to be $3\vec{a}\cdot\vec{b}$, I am not sure why it is that my method fails. I'm fairly confident I may lack some pretty fundamental understanding when it comes to this convention, and if someone would be so kind as to aid my understanding of this convention that would be much appreciated.