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Changed MxMx{0} to \Delta M x {0} like the problem asked for.
Justin Hilburn
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In my experience finding an explicit resolution is rarely possible. Instead you want to learn how to use the six operations. A good reference is section 8.3 of Chriss and Ginzburg.

First a few general facts. Let $X$ be a variety and let $p: X \to pt$ be the unique map to the point. There are two interesting sheaves on $X$: the constant sheaf $k_X = p^* k_{pt}$ and the dualizing sheaf $\mathbb{D}_X = p^! k_{pt}$. Note that the constant sheaf is preserved under $*$-pullbacks and the dualizing sheaf is preserved by $!$-pullbacks. Furthermore if $X$ is smooth we have $\mathbb{D}_X = k_X[\dim_{\mathbb{R}} X]$.

Now we are ready to solve your problem. Let $i: \Delta M \times \{0\} \to M \times M \times \mathbb{R}$. Note that since $i$ is closed we have $i_! = i_*$. Now we compute \begin{align} \mathcal{RHom}(i_! i^* k_{M \times M \times \mathbb{R}}, k_{M \times M \times \mathbb{R}}) &= \mathcal{RHom}(i^* k_{M \times M \times \mathbb{R}}, i^!\mathbb{D}_{M \times M \times \mathbb{R}}[-2m-1]) \\ &= \mathcal{RHom}(k_{\Delta M \times \{0\}}, \mathbb{D}_{\Delta M \times \{0\}}[-2m-1]) \\ &= \mathcal{RHom}(k_{\Delta M \times \{0\}}, k_{\Delta M \times \{0\}}[-m-1]) \\ &= k_{\Delta M \times \{0\}}[-m-1] \end{align} where $m = \dim_{\mathbb{R}} M$.

Justin Hilburn
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