1. Let $T=(1-x^2+3x^3)\left(\frac{5}{2}x^3-\frac{1}{5x^2}\right)^{11}.$ We need the term independent of $x$, i.e. the constant term.

2. General term in the binomial expansion:

For $\left(\frac{5}{2}x^3-\frac{1}{5x^2}\right)^{11},$ the general term is $T_{r+1}=\binom{11}{r}\left(\frac{5}{2}x^3\right)^{11-r}\left(-\frac{1}{5x^2}\right)^r,$ where $r=0,1,2,\dots,11$.

3. Simplify the power of $x$ in the general term:

$x^{3(11-r)}\cdot x^{-2r}=x^{33-3r-2r}=x^{33-5r}.$ So the general term is $T_{r+1}=\binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{33-5r}.$

4. Since the whole expression is multiplied by $(1-x^2+3x^3)$, the constant term can arise in three ways:

• $1 \times$ term with $x^0$ from the binomial part,
• $(-x^2) \times$ term with $x^{-2}$ from the binomial part,
• $(3x^3) \times$ term with $x^{-3}$ from the binomial part.

So we solve:

(i) For $x^0$:

$33-5r=0 \implies r=\frac{33}{5}$ Not an integer, so no such term.

(ii) For $x^{-2}$:

$33-5r=-2 \implies 5r=35 \implies r=7.$ So we need the coefficient of $x^{-2}$ in the binomial part: $\binom{11}{7}\left(\frac{5}{2}\right)^4\left(-\frac{1}{5}\right)^7.$ Now, $\binom{11}{7}=330,$ $\left(\frac{5}{2}\right)^4=\frac{625}{16},$ $\left(-\frac{1}{5}\right)^7=-\frac{1}{78125}.$ Thus coefficient $=330\cdot \frac{625}{16}\cdot\left(-\frac{1}{78125}\right).$ Since $78125=625\cdot 125$, $\frac{625}{78125}=\frac{1}{125}.$ Hence $=330\cdot\left(-\frac{1}{16\cdot 125}\right)=330\cdot\left(-\frac{1}{2000}\right)=-\frac{33}{200}.$

Multiplying by $-x^2$, contribution to constant term is $(-1)\left(-\frac{33}{200}\right)=\frac{33}{200}.$

(iii) For $x^{-3}$:

$33-5r=-3 \implies 5r=36$ Not an integer, so no such term.

5. Therefore, the only contribution to the constant term is $\boxed{\frac{33}{200}}.$

6. Comparing with the stored correct answer: Stored answer is B, i.e. $\frac{33}{200}$, which matches our result.