Well this will be tricky

$R=2(v-1)r/v.n$ where $v$ is the number of edges of the figure, $n$ is the number of edges meeting at point and finally $r$ is the resistance of each edge. This formula works for any polyhedron.

Substituting these values we get $R=(\frac{29}{30}$ ) $\Omega$ = $\boxed{0.97}(\Omega$ )

Describe the icosidodecahedron. The $\frac12$ edgeweights are conductances and are the reciprocal of the $2\Omega$ resistances.

$\text{nvertices}=\text{PolyhedronData}[\text{Icosidodecahedron},\text{VertexCount}];$

$\text{edges}=\text{Table}[e[[1]]\leftrightarrow e[[2]],\{e,\text{PolyhedronData}[\text{Icosidodecahedron},\text{Edges}]\}];$

$g=\text{PlanarGraph}\left[\text{Range}[\text{nvertices}],\text{edges}, \\ \text{VertexLabels}\to \text{Name},\text{EdgeWeight}\to \text{ConstantArray}\left[\frac{1}{2},\text{Length}[\text{edges}]\right]\right];$

The Wolfram Mathematica Mathematica algorithm that does the mesh analysis to compute all vertex to vertex effective resistances.

The PseudoInverse is the Moore-Penrose Inverse .

$\text{resistance}=\text{With}\left[\left\{\Gamma =\text{PseudoInverse}\left[\text{With}\left[\{\text{wam}=\text{WeightedAdjacencyMatrix}[\text{\$\#\$1}]\}, \\ \ \ \text{DiagonalMatrix}\left[\text{Tr}\text{/@}\text{wam}^T\right]-\text{wam}\right]\right]\right\}, \\ \ \ \ \ \text{Outer}[\text{Plus},\text{Diagonal}[\Gamma ],\text{Diagonal}[\Gamma ]]-\Gamma -\Gamma ^T\right]\&;$

The invocation of the algorithm above.

$r=\text{resistance}(g);$

The extraction of the effective resistance across all edges and a tally of the values present (the algorithm outputs resistance for conductance and vis-a-vis).

$\text{result}=\text{Tally}[\text{Table}[r[[e[[1]],e[[2]]]],\{e,\text{edges}\}]] \Longrightarrow \left( \begin{array}{cc} \frac{29}{30} & 60 \\ \end{array} \right)$

This result shows that, in fact, all the effective resistances across each edge are the same. Below are the numeric version of the values.

$N[\text{result}] \Longrightarrow \left( \begin{array}{cc} 0.966667 & 60. \\ \end{array} \right)$