Let varphi(x)=(ln x)/(x) for x>0.

$\varphi'(x)=\frac{1-\ln x}{x^{2}}$, so $\varphi$ increases on $(0,e)$ and decreases on $(e,\infty)$. (A) False: $\varphi(100)>\varphi(101)$ gives $101\ln100>100\ln101$, i.e. $100^{101}>101^{100}$. (B) False: $x=2$ and $x=4$ are solutions, and there is also a negative root near $x\approx-0.767$, since $2^{x}-x^{2}$ is $-\frac12$ at $x=-1$ and $1$ at $x=0$; so there are three real solutions. (C) True: $n^{1/n}>\sqrt2$ means $\varphi(n)>\varphi(2)=\varphi(4)$, which for integers holds only at $n=3$ since $\varphi(1)=0$ and $\varphi$ decreases for $n\ge3$ with $\varphi(4)=\varphi(2)$. (D) False: $\varphi(e)>\varphi(\pi)$ gives $\pi>e\ln\pi$, i.e. $e^{\pi}>\pi^{e}$. Only (C) is correct.