1. Given differential equation

We have $x^2 f'(x)=2xf(x)+3,$ with $f(1)=4.$

We need to find $2f(2)$.

2. Rewrite the equation in standard form

Divide both sides by $x^2$: $f'(x)-\frac{2}{x}f(x)=\frac{3}{x^2}.$

This is a linear differential equation: $f'(x)+P(x)f(x)=Q(x),$ where $P(x)=-\frac{2}{x}, \qquad Q(x)=\frac{3}{x^2}.$

3. Find the integrating factor

The integrating factor is $IF=e^{\int P(x)\,dx}=e^{\int -\frac{2}{x}\,dx}=e^{-2\ln x}=x^{-2},$ since $x>0$.

4. Multiply the equation by the integrating factor

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

Notice that the left side is $\frac{d}{dx}\left(f(x)x^{-2}\right).$

So, $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4}.$

5. Integrate both sides

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

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

Multiplying by $x^2$: $f(x)=Cx^2-\frac{1}{x}.$

6. Use the initial condition

Given $f(1)=4$: $4=C(1)^2-1 \Rightarrow 4=C-1 \Rightarrow C=5.$

Thus, $f(x)=5x^2-\frac{1}{x}.$

7. Compute $f(2)$

$f(2)=5(2)^2-\frac{1}{2}=20-\frac{1}{2}=\frac{39}{2}.$

Therefore, $2f(2)=2\cdot \frac{39}{2}=39.$

8. Check the options

The correct option is $\boxed{39}$ which is Option D.

9. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.