Equation of parabola is $y^2=4 a x$


Let $P$ be the side of the equilateral $\Delta \mathrm{OAB}$ whose one vertex is at the vertex of parabola then by symmetry $A B$ is $\perp$ to the axis $\mathrm{ON}$ of parabola,
Let $\mathrm{ON}=x$, then $\mathrm{BN}=\frac{\mathrm{P}}{2}$
$\therefore(x, \mathrm{P} / 2)$ lies on parabola
$\therefore \quad \frac{\mathrm{P}^2}{4}=4 a x \quad \therefore x=\frac{\mathrm{P}^2}{16 a}$
Now, from $\triangle \mathrm{ONB}, \mathrm{OB}^2=\mathrm{ON}^2+\mathrm{NB}^2$
$\begin{aligned}
&\Rightarrow \mathrm{P}^2=\frac{\mathrm{P}^4}{(16 a)^2}+\frac{\mathrm{P}^2}{4} \\
&\Rightarrow \frac{\mathrm{P}^2}{16^2 a^2}+\frac{1}{4}=1 \Rightarrow \frac{\mathrm{P}^2}{16^2 a^2}=\frac{3}{4}
\end{aligned}$
$\Rightarrow \mathrm{P}^2=16 \times 16 \times a^2 \times \frac{3}{4} \Rightarrow \mathrm{P}=8 \sqrt{3} a$
$\therefore \quad$ Side of triangle is $8 \sqrt{3} a$