= Solution
Expand the propagation distance at small angles:
$$
r\simeq L+\frac{x_0^2+y_0^2}{2L}-\frac{x_0x+y_0y}{L}+\frac{x^2+y^2}{2L}.
$$
The first two terms supply an aperture-independent phase. If the aperture quadratic phase is small, $k(x^2+y^2)/(2L)\ll1$, it too may be dropped from the integrand. The remaining <Fraunhofer diffraction> amplitude is the aperture's <Fourier transform> at transverse wavevector $(kx_0/L,ky_0/L)$, as required.
For slit width $w$ and center spacing $d$, adding the two shifted top-hat transforms gives the normalized intensity
$$
\boxed{\frac{I(X)}{I(0)}=\left[\frac{\sin(\pi wX/(\lambda L))}{\pi wX/(\lambda L)}\right]^2
\cos^2\left(\frac{\pi dX}{\lambda L}\right).}
$$
The aperture outer half-width is $17.5\,\mathrm{\mu m}$; its maximum quadratic phase is only about $1.5\times10^{-3}$ radians at the given wavelength and distance, so the Fraunhofer approximation is excellent. Fringe spacing is $\lambda L/d=25.28\,\mathrm{mm}$ and the first single-slit envelope zero is at $\lambda L/w=63.2\,\mathrm{mm}$.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-double-slit-photon-counting.png]
{title=Two finite-width slits: interference fringes, diffraction envelope and shifted side maximum}
The <side maximum of a finite-width double-slit pattern> shifts slightly inward from $25.28\,\mathrm{mm}$. Differentiating the full intensity gives, away from zeros,
$$
\frac{\pi w}{\lambda L}\left[\cot u-\frac1u\right]
=\frac{\pi d}{\lambda L}\tan v,\quad u=\frac{\pi wX}{\lambda L},\quad v=\frac{\pi dX}{\lambda L}.
$$
Its first side maximum is at $X\simeq23.88\,\mathrm{mm}$ with $I/I(0)\simeq0.592$. The expected rate is therefore $59.2\,\mathrm{s^{-1}}$. Poisson shot noise gives signal-to-noise ratio $N/\sqrt N=\sqrt N$, so one needs $N=100$ counts:
$$
\boxed{t\simeq100/(59.2\,\mathrm{s^{-1}})=1.69\,\mathrm s.}
$$
Using the common fringe-center approximation instead gives relative intensity $\operatorname{sinc}^2(\pi w/d)\simeq0.573$ and $t\simeq1.75\,\mathrm s$. Both express the same order of exposure; the first uses the true maximum. Dark counts, background subtraction and uncertainty in the reference rate are excluded by the ideal stated noise model.
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