So the problem is, given $X\sim N(1,2^2)$, $Y=e^X$, if the cdf of standard normal distribution is $\Phi$, to show that
$$\Phi\left(\frac{\ln Y-1}{2}\right)\sim U[0,1],$$
where $U[0,1]$ is the uniform distribution defined on $[0,1]$.
What I can tell now is that
$$\frac{\ln Y-1}{2}=\frac{X-1}{2}=Z\sim N(0,1^2),$$
is a standard normal distribution. So basically the problem is reduced to showing that $$\Phi(Z)\sim U[0,1].$$
But as you can see in here, the cdf of snd (the red curve) is not "uniform" on $[0,1]$. Rather $N(-2,0.5)$ (the green curve) is. So I'm certainly misunderstanding here something. Please help me to understand.
$\Phi (Z)$ is uniformly distributed on $(0,1)$: $P(\Phi (Z) \leq u)=P(Z \leq \Phi^{-1}(u))=\Phi (\Phi^{-1}(u))=u$ for all $u \in (0,1)$. I have used the fact that $\Phi$ is a continuous strictly increasing function.