The urns A ,   B and C contains 4 red, 6 black; 5 red, 5 black and λ red, 4 black balls…

Let $R$ be the event of drawing red balls.

$P\left(\frac{R}{A}\right)=\frac{4}{10}$

$P\left(\frac{R}{B}\right)=\frac{5}{10}$

$P\left(\frac{R}{C}\right)=\frac{\lambda }{4+\lambda }$

Now,

$P\left(\frac{C}{R}\right)=0.4$

$\Rightarrow \frac{P\left(C\right)P\left({\displaystyle \frac{R}{C}}\right)}{P\left(A\right)P\left(\frac{R}{A}\right)+P\left(B\right)P\left(\frac{R}{B}\right)+P\left(C\right)P\left(\frac{R}{C}\right)}=\frac{4}{10}$

$\Rightarrow \frac{{\displaystyle \frac{1}{3}}\times \left({\displaystyle \frac{\lambda }{\lambda +4}}\right)}{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{4}{10}}+\frac{1}{3}\times \frac{5}{10}+\frac{1}{3}\times \left(\frac{\lambda }{\lambda +4}\right)}=\frac{4}{10}$

$\Rightarrow \frac{{\displaystyle \left(\frac{\lambda }{\lambda +4}\right)}}{{\displaystyle \frac{9}{10}+\left(\frac{\lambda }{\lambda +4}\right)}}=\frac{4}{10}$

$\Rightarrow \frac{\lambda }{\lambda +4}=\frac{4}{10}\left[\frac{9}{10}+\left(\frac{\lambda }{\lambda +4}\right)\right]$

$\Rightarrow \frac{5\lambda }{\lambda +4}=2\left[\frac{9}{10}+\left(\frac{\lambda }{\lambda +4}\right)\right]$

$\Rightarrow \frac{5\lambda }{\lambda +4}=\frac{19\lambda +36}{5\left(\lambda +4\right)}$

$\Rightarrow 25\lambda =19\lambda +36$

$\Rightarrow 6\lambda =36$

$\Rightarrow \lambda =6$

So, parabola is

$y^2=6x=4\times \frac{2}{3}\times x$

Comparing with $y^2=4ax$, we get

$a=\frac{2}{3}$

Hence, we have

Let $P\equiv \left(\frac{2}{3}{t}^2,\frac{4}{3}t\right)$

In $∆OPQ$, we have

$\tan \left(30°\right)=\frac{2at}{a{t}^2}=\frac{2}{t}$

$\Rightarrow \frac{2}{t}=\frac{1}{\sqrt{3}}$

$\Rightarrow t=2\sqrt{3}$

Side length of triangle$=4at=12\sqrt{3}$

Square of side of triangle

$={\left(12\sqrt{3}\right)}^2=432$ units