Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area.
So, $\quad I \propto A$ or $I=k A$
where, $k=$ constant.
Now, if $I_0=$ intensity of parent star.
Then, $I_0=k \pi(100 R)^2=k \pi R^2 \times 10000$
When exoplanet is in front of star,
observed intensity will be minimum. Let intensity minimum is $I_{\min }$, then
$I_{\min }=k\left[\pi(100 R)^2-\pi R^2\right]$
$\Rightarrow I_{\min } =k \pi R^2(10000-1)$
$=k \pi R^2 \times 9999$
$\text { So, } \frac{I_{min}}{I_0} =\frac{k \pi R^2 \times 9999}{k \pi R^2 \times 10000}$
$\Rightarrow I_{\min }=I_0 \times 0.9999$
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