Let $\alpha$ and $\beta$ are the roots of given equation $3m^2{x}^2+m\left(m-4\right)x+2=0$.

$\therefore \alpha +\beta =-\frac{m\left(m-4\right)}{3m^2} \& \alpha \beta =\frac{2}{3m^2} .........\left(i\right)$

Given $\frac{\alpha }{\beta }=\lambda \& \lambda +\frac{1}{\lambda }=1$

$\Rightarrow \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=1$

$\Rightarrow {\left(\alpha +\beta \right)}^2=3\alpha \beta$

Using equation $\left(i\right)$, we get

${\left(\frac{-m\left(m-4\right)}{3m^2}\right)}^2=\frac{3\times 2}{3m^2}$

$\Rightarrow m=4\pm 3\sqrt{2}$

Hence, the least value of $m$ is $4-3\sqrt{2}$.