= Solution
Set $N=(100\delta)^g$ and take $G\sim G(N,4\delta/N)$. Its expected number of edges is asymptotic to $2\delta N$, and a Chernoff bound makes $e(G)>3\delta N/2$ with positive probability. The expected number of cycles of length below $g$ is at most
$$
\sum_{\ell=3}^{g-1}\frac{(4\delta)^\ell}{2\ell}<(4\delta)^g,
$$
which is far below $\delta N/2$; Markov's inequality shows that, simultaneously with positive probability, there are fewer than $\delta N/2$ such cycles.
Choose such a graph and delete one edge from every cycle of length below $g$. The resulting graph has girth at least $g$ and more than $\delta N$ edges, hence average degree greater than $2\delta$. Part (a) supplies a subgraph of minimum degree at least $\delta$. It uses at most $N=(100\delta)^g$ vertices and cannot acquire any shorter cycle.
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