Option B is correct
Substitute $t=\left ( tan\theta \right )^{\frac{1}{3}}$ ------(1)
$\Rightarrow$ $dt=\frac{1}{3}\times(tan \theta)^{\frac{-2}{3}}\times sec^2\theta$ $d\theta$
$\therefore$ The integral would now become $\frac{(tan \theta)^{\frac {2}{3}}\times \frac{1}{3}\times (tan \theta)^{\frac{-2}{3}}\times sec^2\theta}{ 1+tan^2\theta }$ $d\theta$
$\Rightarrow$ $\frac{1}{3}\times$$\int_{0}^{x^2} \theta$ $d\theta= \frac{1}{3}\times$ $\frac{tan^{-1}(x^{6})}{x^{6}}$ (using identity $1+tan^{2}\theta=sec^{2}\theta$ and resubstituting value of $t$ from equation 1)
$\therefore$ limit now becomes $\lim_{x \to0}$ $\frac{1}{3}\times \frac{tan^{-1}(x^{6})}{x^{6}}$
if $x\rightarrow0$ then $x^{6}\rightarrow 0$ . So, using the formula $\lim_{x \to0}$ $\frac{tan^{-1}(x)}{x}=1$, we get the answer as $\frac{1}{3}\times 1=\frac{1}{3}$