The given line passes through the focus $\left(\frac{1}{4},0\right)$, so it is a focal chord for the given parabola.

Hence, its slope is $\tan \alpha =\frac{-1}{\sqrt{3}}$
$\therefore$ the length of the intercept is the same as the length of the focal chord

Hence, the length of the intercept$=4a{\text{cosec}}^2\alpha =4\cdot 1\cdot {\left(2\right)}^2=16$ units