Let $$S(\alpha) = \{(x, y): y^{2} \le x, 0 \le x \le \alpha\}$$ and $$A(\alpha)$$ is area of the region $$S(\alpha)$$. If for a $$\lambda$$, $$0 < \lambda < 4$$, $$A(\lambda):A(4) = 2:5$$, then $$\lambda$$ equals:

We have the set $$S(\alpha)=\{(x,y):y^{2}\le x,\;0\le x\le\alpha\}.$$

Geometrically, the inequality $$y^{2}\le x$$ represents the region to the right of the parabola $$x=y^{2}$$, while $$0\le x\le\alpha$$ restricts us to the vertical strip beginning at the y-axis and ending at the line $$x=\alpha$$. Hence, for every admissible value of $$y$$, the variable $$x$$ starts from the parabola $$x=y^{2}$$ and goes up to $$x=\alpha$$.

To find the area $$A(\alpha)$$ of this region, we use the definition of area in terms of a double integral:

$$ A(\alpha)=\iint_{S(\alpha)}\!dx\,dy. $$

For a fixed $$y$$, $$x$$ varies from $$y^{2}$$ to $$\alpha$$, and $$y$$ itself can vary as long as $$y^{2}\le\alpha$$, i.e. $$-\sqrt{\alpha}\le y\le\sqrt{\alpha}$$. Therefore, we may write

$$ A(\alpha)=\int_{y=-\sqrt{\alpha}}^{\sqrt{\alpha}} \!\left(\int_{x=y^{2}}^{\alpha}dx\right)dy. $$

Evaluating the inner integral first, we obtain

$$ \int_{x=y^{2}}^{\alpha}dx=\alpha-y^{2}. $$

This gives

$$ A(\alpha)=\int_{y=-\sqrt{\alpha}}^{\sqrt{\alpha}}\!(\alpha-y^{2})\,dy. $$

The integrand $$\alpha-y^{2}$$ is an even function of $$y$$, so we use symmetry about the x-axis:

$$ A(\alpha)=2\int_{0}^{\sqrt{\alpha}}(\alpha-y^{2})\,dy. $$

Now we integrate term by term:

$$ \int_{0}^{\sqrt{\alpha}}\alpha\,dy=\alpha\,y\Big|_{0}^{\sqrt{\alpha}} =\alpha\sqrt{\alpha}, $$

and

$$ \int_{0}^{\sqrt{\alpha}}y^{2}\,dy=\frac{y^{3}}{3}\Big|_{0}^{\sqrt{\alpha}} =\frac{(\sqrt{\alpha})^{3}}{3} =\frac{\alpha^{3/2}}{3}. $$

Substituting these results,

$$ A(\alpha)=2\left(\alpha\sqrt{\alpha}-\frac{\alpha^{3/2}}{3}\right) =2\alpha^{3/2}\left(1-\frac{1}{3}\right) =2\alpha^{3/2}\left(\frac{2}{3}\right) =\frac{4}{3}\,\alpha^{3/2}. $$

Thus

$$ A(\alpha)=\frac{4}{3}\,\alpha^{3/2}. $$

We are told that for some $$\lambda$$ with $$0<\lambda<4$$,

$$ \frac{A(\lambda)}{A(4)}=\frac{2}{5}. $$

Using the general formula, we find

$$ A(4)=\frac{4}{3}\,4^{3/2}. $$

Because $$4^{3/2}=(\sqrt{4})^{3}=2^{3}=8$$, this simplifies to

$$ A(4)=\frac{4}{3}\times8=\frac{32}{3}. $$

Now we express $$A(\lambda)$$ in terms of $$\lambda$$:

$$ A(\lambda)=\frac{4}{3}\,\lambda^{3/2}. $$

The given ratio becomes

$$ \frac{\dfrac{4}{3}\lambda^{3/2}}{\dfrac{32}{3}}=\frac{2}{5}. $$

Multiplying numerator and denominator by $$3$$ to clear the fractions, we get

$$ \frac{4\lambda^{3/2}}{32}=\frac{2}{5}. $$

Simplifying the left side by dividing numerator and denominator by $$4$$,

$$ \frac{\lambda^{3/2}}{8}=\frac{2}{5}. $$

Now we solve for $$\lambda^{3/2}$$:

$$ \lambda^{3/2}=8\left(\frac{2}{5}\right)=\frac{16}{5}. $$

To obtain $$\lambda$$ itself, we raise both sides to the power $$\tfrac{2}{3}$$ (the inverse of $$\tfrac{3}{2}$$):

$$ \lambda=\left(\frac{16}{5}\right)^{2/3}. $$

Because $$16=4^{2}$$, we can rewrite the numerator inside the radical to match the options more clearly:

$$ \left(\frac{16}{5}\right)^{2/3} =\left(\frac{4^{2}}{5}\right)^{2/3} =\left(\frac{4^{2}}{5^{2}}\right)^{1/3} =\left(\frac{4}{25}\right)^{1/3}\!\!\cdot4. $$

Thus

$$ \lambda=4\left(\frac{4}{25}\right)^{1/3}. $$

This expression coincides exactly with Option C.

Hence, the correct answer is Option C.