Euler's Theorem is not needed. It can be completely solved using only the Binomial Theorem:
$$\rm 9^{\color{#c00}{\large 10}} =\ (-1+10)^{\color{#c00}{\large 10}} =\: (-1)^{\color{#c00}{\large 10}} - \color{#c00}{10}\cdot 10 + 10^{\large 2}\:(\cdots)\ \color{}{\equiv\ 1}\ \ (mod\ 100)$$
So $\rm \bmod 100\!:\, \ 9^{\large 9^{\LARGE 9}}\!\!\equiv\ 9^{\large 9^{\LARGE 9}\, mod\ \color{#c00}{10}} \equiv\ 9^{\large (-1)^{\LARGE 9}}\!\! \equiv 9^{\large -1}\!\equiv \dfrac{1}9 \equiv \dfrac{-99}9 \equiv {-}11 \equiv 89 $
Remark $ $ Above we used the useful fact that if the powers of $\,a=9\,$ repeat with period length $\color{#c00}{10}\,$ then all exponents on $\,a\,$ can be taken modulo $\,\color{#c00}{10}.\,$ Said more precisely we used the following
$$\ \ \color{#c00}{a^{\color{#c00}{\large 10}}\equiv 1}\!\!\pmod{\!m},\,\ J\equiv K\!\!\!\pmod{\!\color{#c00}{10}}\ \,\Rightarrow\,\ a^{\large J}\equiv a^{\large K}\!\!\!\!\pmod{\!m}$$
for the specific values $\ a=9,\,$ and $\,J = 9^{\large 9},\,$ and $\,K = (9^{\large 9}\,{\rm mod}\ 10).\,$ A proof is easy:
$$ J = K\! +\! 10N\,\Rightarrow\, a^{\large J}\! = a^{\large K+10N}\! = a^{\large K} (\color{#c00}{\large a^{10}})^{\large N}\!\equiv a^{\large K} \color{#c00}1^{\large N}\!\equiv a^{\large K}\!\!\!\!\pmod{\!m}\qquad $$
where we have employed the $ $ Congruence Product and Power Rules. For further discussion see modular order reduction.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
An alternative approach, exploiting the Chinese remainder theorem and the fact that $2$ is a generator in $\mathbb{Z}/(25\mathbb{Z})^*$:
$$ 17^{198}\equiv (-8)^{198} \equiv 2^{3\cdot 198} \equiv 2^{14} \equiv 16\cdot 1024\equiv -16\equiv 9\pmod{25} $$
$$ 17^{198}\equiv 1\pmod{4} $$
together imply $17^{198}\equiv \color{red}{9}\pmod{100}$.
Best Answer
$2^{20}=1048576\equiv1\pmod{25},$ so $2^{400}\equiv1\pmod{25},$ so $25$ divides $2^{400}-1,$
so $100$ divides $2^{402}-4,$ so $2^{402}\equiv4 \pmod {100}.$ Can you take it from here?