$\newcommand\ep{\epsilon}$$\newcommand\de{\delta}$We shall be assuming that $\ep\in(0,1/e]$. Note that $l(x):=(\ln x)/x$ is decreasing in $x\ge e$. 
So, for $x\ge e$ we have 
$$(\ln x)/x\le\ep\iff x\ge x_\ep,$$ 
where $x_\ep\in[e,\infty)$ is the root of the equation
$$l(x_\ep)=\ep.$$
Letting 
$$y:=y_\ep:=\frac1\ep\,\ln\frac1\ep\ge e,$$
we have 
$$l(y)=\ep\frac{\ln\frac1\ep+\ln\ln\frac1\ep}{\ln\frac1\ep}\ge\ep=l(x_\ep)$$
and hence 
$$x_\ep\ge y=\frac1\ep\,\ln\frac1\ep.\tag{1}$$

On the other hand, for each real $\de>0$, letting 
$$z:=z_\ep:=(1+\de)y_\ep$$
we have 
$$l(z)=\ep\frac{\ln\frac{1+\de}\ep+\ln\ln\frac1\ep}{(1+\de)\ln\frac1\ep}\le\ep=l(x_\ep)$$
for all small enough $\ep>0$ and hence 
$$x_\ep\le z=\frac{1+\de}\ep\,\ln\frac1\ep.\tag{2}$$

In particular, it follows that 
$$x_\ep\sim\frac1\ep\,\ln\frac1\ep$$
as $\ep\downarrow0$. 

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Working similarly but just a bit harder, we can see that the upper bound in (2) on $x_\ep$ will hold if, instead of taking a constant $\de>0$, we take 
$$\de=c\eta,\quad\text{with}\quad \eta:=\frac{\ln\ln\frac1\ep}{\ln\frac1\ep},$$
for any fixed real $c>1$ and all small enough $\ep>0$. 
On the other hand, the lower bound on $x_\ep$ in (1) can be improved as follows: 
$$x_\ep\ge \frac{1+\eta}\ep\,\ln\frac1\ep$$
for all real $x\ge e$.