$\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$$ as $\ep\downarrow0$.
Working similarly but just a bit harder, we can see that the upper bound in (2) on $x_\ep$ will hold if, instead of taking a constant $\de>0$, we take $$\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 $x\ge e$.