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

$I_{\min }=k\left[\pi(100 R)^2-\pi R^2\right]$

$\Rightarrow I_{\min } =k \pi R^2(10000-1)$

$=k \pi R^2 \times 9999$

$\text { So, } \frac{I_{min}}{I_0} =\frac{k \pi R^2 \times 9999}{k \pi R^2 \times 10000}$

$\Rightarrow I_{\min }=I_0 \times 0.9999$