= Solution
For two identical, head-on beams with <Gaussian distributions>, integrating the product of their normalized transverse densities gives an overlap $1/(4\pi\sigma_x\sigma_y)$. Each of the $n_b$ colliding bunch pairs crosses once per revolution, with $f_{\rm rev}\simeq c/C$. Thus
$$
\mathcal L=\frac{f_{\rm rev}n_bN_1N_2}{4\pi\sigma_x\sigma_y}=\frac{(2.998\times10^8/2.7\times10^4)\,2800\,(1.15\times10^{11})^2}{4\pi(2\times10^{-5})^2}.
$$
Therefore
$$
\boxed{\mathcal L\simeq8.18\times10^{37}\ \mathrm{m}^{-2}\mathrm{s}^{-1}=8.18\times10^{33}\ \mathrm{cm}^{-2}\mathrm{s}^{-1}}.
$$
The factor $4\pi$ accounts for the convolution of two beams, rather than the $2\pi$ normalization of just one beam. The result assumes all stated bunch pairs collide at the chosen interaction point, speeds approximately $c$, zero crossing angle and negligible hourglass reduction. Additional interaction points do not multiply the local <collider luminosity> at a single detector.
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