= Solution
The incident intensity radius is $w_L=2.5\,\mathrm{mm}$, half the quoted diameter. For an ideal collimated <Gaussian beam> focused by a thin <lens>, the diffraction-limited waist is
$$
w_0\simeq\frac{f\lambda_0}{\pi w_L}=2.037\,\mu\mathrm m.
$$
The incident <Rayleigh range> $\pi w_L^2/\lambda_0\simeq24.5\,\mathrm m$ is much larger than $f$, justifying this approximation. The pulse <energy> is $U=P_{\rm av}/R=0.25\,\mu\mathrm J$. Taking its duration as an effective energy-to-peak-power width gives $P_{\rm pk}\simeq U/\tau=7.14\times10^5\,\mathrm W$. Since $I(r)=I_0e^{-2r^2/w_0^2}$ has integrated power $I_0\pi w_0^2/2$,
$$
\boxed{I_{\rm pk}\simeq\frac{2U}{\pi w_0^2\tau}=1.10\times10^{17}\,\mathrm{W\,m^{-2}}=1.10\times10^{13}\,\mathrm{W\,cm^{-2}}.}
$$
If $350\,\mathrm{fs}$ denotes the temporal intensity full width at half maximum of a <Gaussian pulse>, the time integral instead gives $P_{\rm pk}=2\sqrt{\ln2}\,U/(\sqrt\pi\tau)$, and $I_{\rm pk}=1.03\times10^{13}\,\mathrm{W\,cm^{-2}}$. The pulse shape is unspecified, so the estimate's order-one convention is explicit. Aberration, clipping and propagation through glass were excluded from this calculation as stipulated.
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