Let A be the area of contact of each tyre with the ground, when the person rides the bicycle.

Then, pressure exerted by the person on the each tyre,

$\mathrm{P}=\frac{\mathrm{weight}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{person}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{bicycle}}{\mathrm{area}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{contact}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{tyres}}=\frac{\mathrm{Mg}}{2\times \mathrm{A}}$

$=\frac{90\times 9.8}{2\times \mathrm{A}}=\frac{441}{\mathrm{A}}$

Since the gauge pressure in both the tyres of the bicycle is $6.9\times {10}^5 \mathrm{Pa}$, we have

$\frac{441}{\mathrm{A}}=6.9\times {10}^5$

Or $\mathrm{A}=\frac{441}{6.9\times {10}^5}=6.39\times {10}^{-4} {\mathrm{m}}^2$