1. Use Bayes' theorem to find $\lambda$

Let $R$ be the event that the drawn ball is red.

Since one of the three urns is selected at random, $P(A)=P(B)=P(C)=\frac13.$

Red-ball probabilities from each urn:

• Urn $A$: $P(R|A)=\frac{4}{10}=\frac25$
• Urn $B$: $P(R|B)=\frac{5}{10}=\frac12$
• Urn $C$: $P(R|C)=\frac{\lambda}{\lambda+4}$

Given: $P(C|R)=0.4=\frac25.$

By Bayes' theorem, $P(C|R)=\frac{P(C)P(R|C)}{P(A)P(R|A)+P(B)P(R|B)+P(C)P(R|C)}.$

Since all urns are equally likely, $\frac13$ cancels: $\frac{\lambda}{\lambda+4}\Bigg/\left(\frac25+\frac12+\frac{\lambda}{\lambda+4}\right)=\frac25.$

Now, $\frac25+\frac12=\frac{4+5}{10}=\frac9{10}.$

So, $\frac{\lambda}{\lambda+4}\Bigg/\left(\frac9{10}+\frac{\lambda}{\lambda+4}\right)=\frac25.$

Let $t=\frac{\lambda}{\lambda+4}.$ Then $\frac{t}{\frac9{10}+t}=\frac25.$

Cross-multiplying: $5t=2\left(\frac9{10}+t\right)=\frac95+2t$ $3t=\frac95$ $t=\frac35.$

Hence, $\frac{\lambda}{\lambda+4}=\frac35$ $5\lambda=3\lambda+12$ $2\lambda=12$ $\lambda=6.$

2. Parabola becomes $y^2=6x.$

Compare with standard form $y^2=4ax$: $4a=6\implies a=\frac32.$

So the parabola is $y^2=4ax \quad \text{with } a=\frac32.$

3. Coordinates of an equilateral triangle with one vertex at the parabola vertex

Vertex of parabola is at $O=(0,0).$

Take the other two vertices on the parabola as $P(at_1^2,2at_1), \qquad Q(at_2^2,2at_2).$

Since triangle $OPQ$ is equilateral, we need $OP=OQ=PQ.$

First, $OP^2=(at_1^2)^2+(2at_1)^2=a^2t_1^4+4a^2t_1^2=a^2t_1^2(t_1^2+4).$ Similarly, $OQ^2=a^2t_2^2(t_2^2+4).$

A standard symmetric choice for the largest such triangle is to take points symmetric about the $x$-axis: $P(at^2,2at), \qquad Q(at^2,-2at).$

Then $PQ=4at,$ and $OP^2=OQ^2=a^2t^2(t^2+4).$

For equilateral triangle, $OP=PQ.$ So, $a^2t^2(t^2+4)=(4at)^2=16a^2t^2.$

Assuming $t\ne 0$, divide by $a^2t^2$: $t^2+4=16$ $t^2=12.$

Thus side length squared is $PQ^2=(4at)^2=16a^2t^2.$ Substitute $a=\frac32$ and $t^2=12$: $PQ^2=16\left(\frac32\right)^2(12)=16\cdot\frac94\cdot 12.$

Now, $16\cdot\frac94=36,$ so $PQ^2=36\cdot 12=432.$

4. Final answer

The square of the side of the largest equilateral triangle is $\boxed{432}.$