Given that,
$ 4^{n} > 17^{19} $
Taking the $ \log_{10}$ both sides.
$ \log_{10}4^{n} > \log_{10} 17^{19} $
$ \Rightarrow n \log_{10}4 > 19 \log_{10}17 \quad [ \because \log_{b}a^{x} = x \log_{b}a] $
$ \Rightarrow n > 19 \left( \frac{\log_{10}17}{ \log_{10}4} \right) $
$ \Rightarrow n > 19 ( \log_{4}17) \quad \left[ \because \log_{b}a = \frac{ \log_{x}a}{ \log_{x}b} \right] $
For easy calculation, let us assume $17 \approx 16.$
Now, $ n > 19 ( \log_{4} 16)$
$ \Rightarrow n > 19 ( \log_{4} 4^{2})$
$ \Rightarrow n > 19 \times 2 ( \log_{4}4)$
$ \Rightarrow \boxed{n >38} \quad [ \because \log_{a}a = 1 ] $
$ \Rightarrow \boxed{n \simeq 39} $
$\therefore$ The small integer $n$ is closed to $39.$
$\textbf{Short Method :}$
Given that, $ 4^{n} > 17^{19} $
$ \Rightarrow \left( 4^{2} \right)^{\frac{n}{2}} > 17^{19} $
$ \Rightarrow (16)^{\frac{n}{2}} > 17^{19} $
Here, $ 16 < 17 ,$ so $ \frac{n}{2}$ must be greater than $19.$
Thus, $ \frac{n}{2} > 19 $
$ \Rightarrow n > 38 $
$ \Rightarrow \boxed{n \simeq 39} $
Correct Answer $:\text{C}$