1. Given equation

We have $x^2 f(x) - x = 4\int_0^x t f(t)\,dt$ and $f(1)=\frac23.$

We need to find $18f(3)$.

2. Differentiate both sides

Differentiate with respect to $x$:

Left side: $\frac{d}{dx}\big(x^2 f(x)-x\big)=2x f(x)+x^2 f'(x)-1.$

Right side: By the Fundamental Theorem of Calculus, $\frac{d}{dx}\left(4\int_0^x t f(t)\,dt\right)=4x f(x).$

So, $2x f(x)+x^2 f'(x)-1=4x f(x).$

Rearrange: $x^2 f'(x)-2x f(x)-1=0,$ that is, $x^2 f'(x)-2x f(x)=1.$

For $x\neq 0$, $f'(x)-\frac{2}{x}f(x)=\frac{1}{x^2}.$

3. Solve the linear differential equation

The differential equation is $f'(x)-\frac{2}{x}f(x)=\frac{1}{x^2}.$

Its integrating factor is $\text{I.F.}=e^{\int -2/x\,dx}=e^{-2\ln x}=x^{-2}$ (for $x>0$, which is enough since we use $x=1,3$).

Multiply the equation by $x^{-2}$: $x^{-2}f'(x)-2x^{-3}f(x)=x^{-4}.$

Recognize the left side as a derivative: $\frac{d}{dx}\left(f(x)x^{-2}\right)=x^{-4}.$

Integrate: $f(x)x^{-2}=\int x^{-4}\,dx=-\frac{1}{3}x^{-3}+C.$

Hence, $f(x)=x^2\left(-\frac{1}{3}x^{-3}+C\right)=-\frac{1}{3x}+Cx^2.$

4. Use the condition $f(1)=\frac23$

Substitute $x=1$: $\frac23=f(1)=-\frac13+C.$

So, $C=1.$

Therefore, $f(x)=x^2-\frac{1}{3x}.$

5. Find $f(3)$

$f(3)=3^2-\frac{1}{3\cdot 3}=9-\frac19=\frac{80}{9}.$

Then, $18f(3)=18\cdot \frac{80}{9}=2\cdot 80=160.$

6. Check options

The value is $18f(3)=160.$

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.