1. Use Vieta's formulas

For the quadratic $x^2-3x+p=0,$ with roots $\alpha,\beta$: $\alpha+\beta=3, \qquad \alpha\beta=p.$

For the quadratic $x^2-6x+q=0,$ with roots $\gamma,\delta$: $\gamma+\delta=6, \qquad \gamma\delta=q.$

2. Use the condition that $\alpha,\beta,\gamma,\delta$ are in geometric progression

Let the four numbers in GP be $a, ar, ar^2, ar^3.$ Then we may take $\alpha=a,\quad \beta=ar,\quad \gamma=ar^2,\quad \delta=ar^3.$

So, $\alpha+\beta=a(1+r)=3 \quad \cdots (1)$ $\gamma+\delta=ar^2(1+r)=6 \quad \cdots (2)$

Divide (2) by (1): $\frac{ar^2(1+r)}{a(1+r)}=\frac{6}{3}$ $r^2=2.$

Hence, $r=\pm \sqrt{2}.$ But for our required ratio, only $r^2=2$ is needed.

3. Find $a$ using (1)

From $a(1+r)=3,$ we get $a=\frac{3}{1+r}.$

4. Compute $p$ and $q$

Since $p=\alpha\beta=a\cdot ar=a^2r,$ we get $p=\left(\frac{3}{1+r}\right)^2 r=\frac{9r}{(1+r)^2}.$

Also, $q=\gamma\delta=(ar^2)(ar^3)=a^2r^5,$ so $q=\left(\frac{3}{1+r}\right)^2 r^5=\frac{9r^5}{(1+r)^2}.$

Now use $r^2=2$: $r^4=4 \Rightarrow r^5=4r.$ Thus, $q=\frac{9\cdot 4r}{(1+r)^2}=4p.$

5. Find the required ratio

We need $(2q+p):(2q-p).$ Since $q=4p$, $2q+p=2(4p)+p=9p,$ $2q-p=2(4p)-p=7p.$

Therefore, $(2q+p):(2q-p)=9p:7p=9:7.$

6. Check options

The correct option is: $\boxed{9:7}$ which is Option A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.