1. We are given $\int_0^x t f(t)\,dt = x^2 f(x), \quad x>0$ and also $f(2)=3.$ We need to find $f(6)$.

2. Differentiate both sides with respect to $x$.

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

Using the product rule on the right: $\frac{d}{dx}\left(x^2 f(x)\right)=2x f(x)+x^2 f'(x).$

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

3. Rearrange: $0=x f(x)+x^2 f'(x)$ $x^2 f'(x)=-x f(x).$ Since $x>0$, divide by $x$: $x f'(x)=-f(x).$

Thus, $f'(x)=-\frac{f(x)}{x}.$ This is a separable differential equation.

4. Solve the differential equation: $\frac{f'(x)}{f(x)}=-\frac{1}{x}.$ Integrate: $\int \frac{f'(x)}{f(x)}\,dx=\int -\frac{1}{x}\,dx$ $\ln |f(x)|=-\ln x + C.$ Hence, $f(x)=\frac{C}{x}$ for some constant $C$.

5. Use the condition $f(2)=3$: $3=\frac{C}{2} \Rightarrow C=6.$ So, $f(x)=\frac{6}{x}.$

6. Now compute $f(6)$: $f(6)=\frac{6}{6}=1.$

7. Check with the integral equation: $\int_0^x t f(t)\,dt=\int_0^x t\cdot \frac{6}{t}\,dt=\int_0^x 6\,dt=6x,$ and $x^2 f(x)=x^2\cdot \frac{6}{x}=6x.$ So the solution is consistent.

Therefore, $f(6)=1.$ So the correct option is A.