1. Let the roots of $x^2+(2-\lambda)x+(10-\lambda)=0$ be $\alpha,\beta$.

Then by Vieta's formulas, $\alpha+\beta=\lambda-2, \qquad \alpha\beta=10-\lambda.$

2. We need to minimize the sum of cubes of the roots: $\alpha^3+\beta^3.$

Use the identity $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).$

Substitute: $\alpha^3+\beta^3=(\lambda-2)^3-3(10-\lambda)(\lambda-2).$

Factor out $(\lambda-2)$: $\alpha^3+\beta^3=(\lambda-2)\left[(\lambda-2)^2-3(10-\lambda)\right].$

Now simplify: $(\lambda-2)^2=\lambda^2-4\lambda+4$ and $-3(10-\lambda)=-30+3\lambda.$

So, $\alpha^3+\beta^3=(\lambda-2)(\lambda^2-\lambda-26).$

Expanding, $S(\lambda)=\lambda^3-3\lambda^2-24\lambda+52.$

3. To find the minimum of $S(\lambda)$, differentiate: $S'(\lambda)=3\lambda^2-6\lambda-24=3(\lambda^2-2\lambda-8)=3(\lambda-4)(\lambda+2).$

Critical points are: $\lambda=4,\,-2.$

Second derivative: $S''(\lambda)=6\lambda-6.$

At $\lambda=4$, $S''(4)=18>0,$ so $\lambda=4$ gives a local minimum.

4. Now the equation becomes, for $\lambda=4$: $x^2+(2-4)x+(10-4)=0$ $x^2-2x+6=0.$

If roots are $r_1,r_2$, then $|r_1-r_2|=\sqrt{(r_1+r_2)^2-4r_1r_2}$ for real roots up to sign, and in magnitude generally from the discriminant, $|r_1-r_2|=\sqrt{|D|}$ with $D=b^2-4ac=(-2)^2-4(1)(6)=4-24=-20.$

Thus the magnitude of the difference is $|r_1-r_2|=\sqrt{20}=2\sqrt{5}.$

5. Hence the correct option is $\boxed{2\sqrt{5}}.$