From the number line shown, we see that $x<x^{3}<x^{2}$.
If $x>1$, then successive powers of $x$ are increasing (that is, $x<x^{2}<x^{3}$ ).
Since this is not the case, then it is not true that $x>1$.
If $x=0$ or $x=1$, then successive powers of $x$ are equal. This is not the case either.
If $0<x<1$, then successive powers of $x$ are decreasing (that is, $x^{3}<x^{2}<x$ ). This is not the case either.
Therefore, it must be the case that $x<0$.
If $x<-1$, we would have $x^{3}<x<0<x^{2}$. This is because when $x<-1$, then $x$ is negative and we have $x^{2}>1$ which gives $x^{3}=x^{2} \times x<1 \times x$. This is not the case here either.
Therefore, it must be the case that $-1<x<0$.
From the given possibilities, this means that $-\frac{2}{5}$ is the only possible value of $x$.
We can check that if $x=-\frac{2}{5}=-0.4$, then $x^{2}=0.16$ and $x^{3}=-0.064$, and so we have $x<x^{3}<x^{2}$. We can also check by substitution that none of the other possible answers gives the correct ordering of $x, x^{2}$ and $x^{3}$.