The positive value of $$\lambda$$ for which the co-efficient of $$x^2$$ in the expansion $$x^2\left(\sqrt{x} + \frac{\lambda}{x^2}\right)^{10}$$ is 720, is:

First we rewrite the given expression in a form that makes every power of $$x$$ visible. We have

$$x^{2}\Bigl(\sqrt{x}+\dfrac{\lambda}{x^{2}}\Bigr)^{10} = x^{2}\Bigl(x^{1/2}+ \lambda\,x^{-2}\Bigr)^{10}.$$

To expand $$(A+B)^{10}$$ we use the Binomial Theorem, stated as $$ (A+B)^{10}= \sum_{r=0}^{10} {10\choose r} A^{\,10-r} B^{\,r}. $$ Here $$A = x^{1/2},\qquad B = \lambda\,x^{-2}.$$

So each general term of the expansion of $$\bigl(x^{1/2}+ \lambda\,x^{-2}\bigr)^{10}$$ is

$$ {10\choose r}\,(x^{1/2})^{\,10-r}\,(\lambda\,x^{-2})^{\,r} = {10\choose r}\,\lambda^{\,r}\,x^{\,(10-r)\cdot\frac12}\,x^{-2r}. $$

Simplifying the exponents of $$x$$ inside that term, we multiply the two powers of $$x$$: $$(10-r)\cdot\frac12 = \frac{10-r}{2},$$ so the total power of $$x$$ in the term (before multiplying by the outer $$x^{2}$$) is

$$\frac{10-r}{2}\;+\;(-2r)=\frac{10-r-4r}{2}=\frac{10-5r}{2}.$$

Now we multiply the whole expansion by the prefactor $$x^{2}$$ that sits outside the bracket. Hence, after multiplication, the power of $$x$$ in the general term becomes

$$ 2+\frac{10-5r}{2}= \frac{4+10-5r}{2}= \frac{14-5r}{2}. $$

We need the coefficient of $$x^{2}$$, so we set this power equal to 2:

$$ \frac{14-5r}{2}=2 \;\Longrightarrow\; 14-5r=4 \;\Longrightarrow\; 5r = 10 \;\Longrightarrow\; r = 2. $$

Thus only the term with $$r=2$$ contributes to the $$x^{2}$$ term.

Substituting $$r=2$$ back into the coefficient part of the term, we get

Coefficient $$= {10\choose 2}\,\lambda^{2} = \frac{10\times9}{2}\,\lambda^{2}=45\,\lambda^{2}.$$

The problem states that this coefficient equals 720, so

$$45\,\lambda^{2}=720 \;\Longrightarrow\; \lambda^{2}= \frac{720}{45}=16 \;\Longrightarrow\; \lambda = 4,$$

and we keep the positive value because the question asks for a positive $$\lambda$$.

Hence, the correct answer is Option C.