1. We need the constant term in $\left(3x^3-2x^2+\frac{5}{x^5}\right)^{10}.$

2. In the general term of the multinomial expansion, if we choose:

• $a$ times the term $3x^3$,
• $b$ times the term $-2x^2$,
• $c$ times the term $5x^{-5}$,

then $a+b+c=10$ and the corresponding term is $\frac{10!}{a!b!c!}(3x^3)^a(-2x^2)^b(5x^{-5})^c.$

3. For this term to be constant, the power of $x$ must be zero: $3a+2b-5c=0.$

So we solve $a+b+c=10$ $3a+2b-5c=0.$

4. Substitute $c=10-a-b$ into the second equation: $3a+2b-5(10-a-b)=0$ $3a+2b-50+5a+5b=0$ $8a+7b=50.$

Now find non-negative integer solutions.

Testing values of $a$:

• If $a=1$, then $8+7b=50\Rightarrow 7b=42\Rightarrow b=6$. Then $c=10-1-6=3.$

This gives the valid solution $(a,b,c)=(1,6,3).$

No other non-negative integer solution works.

5. Therefore the constant term is $\frac{10!}{1!6!3!}(3)^1(-2)^6(5)^3.$

Now compute: $\frac{10!}{1!6!3!}=\frac{10\cdot 9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=840,$ $(-2)^6=64.$

Hence constant term $=840\cdot 3\cdot 64\cdot 125.$

First, $64\cdot 125=8000.$ So $840\cdot 3\cdot 8000=2520\cdot 8000=20,160,000.$

6. We are told this constant term is of the form $2^k\cdot l$, where $l$ is odd.

Now factorize: $20,160,000=2520\cdot 8000.$ Also, $2520=2^3\cdot 3^2\cdot 5\cdot 7,$ $8000=2^6\cdot 5^3.$ So, $20,160,000=2^{3+6}\cdot (3^2\cdot 5^4\cdot 7)=2^9\cdot l,$ where $l$ is odd.

Thus, $k=9.$

7. Therefore the correct option is $\boxed{9}.$