$R =\left\{( a , b ): a \leq b ^{3}\right\}$

It is observed that $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R ,$ since, $\frac{1}{2}>\left(\frac{1}{2}\right)^{3}$

$\therefore R$ is not reflexive.

Now, $(1,2)\in R($ as $1<2^{3}=8)$

But, $(2,1)\notin R$ $($ as $2^{3}>1$ $)$

$\therefore R$ is not symmetric.

We have $\left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \in R,$

since $3<\left(\frac{3}{2}\right)^{2}$ and $\frac{3}{2}<\left(\frac{6}{5}\right)^{3}$

But $\left(3, \frac{6}{5}\right) \notin R$ as $3>\left(\frac{6}{5}\right)^{3}$

$\therefore R$ is not transitive.

Hence, $R$ is neither reflexive, nor symmetric, nor transitive.