1. Let the two roots of $\left(a^2-5a+3\right)x^2+(3a-1)x+2=0$ be $r$ and $2r$.

Then, by Vieta's formulas for $Ax^2+Bx+C=0,$ we have: $r+2r=3r=-\frac{B}{A},$ $r\cdot 2r=2r^2=\frac{C}{A},$ where $A=a^2-5a+3,\quad B=3a-1,\quad C=2.$

So, $3r=-\frac{3a-1}{a^2-5a+3}$ and $2r^2=\frac{2}{a^2-5a+3}.$ Hence, $r^2=\frac{1}{a^2-5a+3}.$

2. Now square the expression for $r$ obtained from the sum of roots.

From $3r=-\frac{3a-1}{a^2-5a+3},$ we get $r=-\frac{3a-1}{3(a^2-5a+3)}.$ Therefore, $r^2=\frac{(3a-1)^2}{9(a^2-5a+3)^2}.$

But also, $r^2=\frac{1}{a^2-5a+3}.$ So, $\frac{(3a-1)^2}{9(a^2-5a+3)^2}=\frac{1}{a^2-5a+3}.$

Multiplying through, $(3a-1)^2=9(a^2-5a+3).$

3. Expand both sides: $9a^2-6a+1=9a^2-45a+27.$

Cancel $9a^2$ from both sides: $-6a+1=-45a+27.$

So, $39a=26,$ $a=\frac{26}{39}=\frac{2}{3}.$

4. Check that the quadratic remains quadratic: $A=a^2-5a+3=\frac{4}{9}-\frac{10}{3}+3=\frac{4-30+27}{9}=\frac{1}{9}\ne 0.$ So this value is valid.

5. Compare with options:

• A: $-\frac13$
• B: $\frac23$
• C: $-\frac23$
• D: $\frac13$

Thus the correct option is $\boxed{\frac23}$ which is option $\boxed{\text{B}}$.