1. Use the arithmetic progression condition

If $a,b,c$ are in A.P., then $2b=a+c.$ Also, the arithmetic mean of $a,b,c$ is $8$, so $\frac{a+b+c}{3}=8 \implies a+b+c=24.$

For three numbers in A.P., the middle term equals the average of all three terms, so $b=8.$ Then from $a+b+c=24$, $a+c=16.$

Since $a,b,c$ are in A.P. with middle term $8$, $a=8-d,\quad b=8,\quad c=8+d$ for some real $d$.

2. Use the geometric progression condition

Given $a+1,\ b,\ c+3$ are in G.P., so $b^2=(a+1)(c+3).$ Since $b=8$, $64=(a+1)(c+3).$

Now use $c=16-a$: $64=(a+1)(19-a).$ Expanding, $64=-a^2+18a+19.$ So $a^2-18a+45=0.$

Solve: $a=\frac{18\pm\sqrt{324-180}}{2}=\frac{18\pm 12}{2}.$ Hence, $a=15 \quad \text{or} \quad a=3.$

Given $a>10$, we take $a=15.$ Then $c=16-a=1, \qquad b=8.$

3. Find the geometric mean of $a,b,c$

The geometric mean $G$ of $a,b,c$ is $G=(abc)^{1/3}.$ Therefore, $G^3=abc.$ Now, $abc=15\cdot 8\cdot 1=120.$

4. Compare with options

Thus, the cube of the geometric mean is $\boxed{120}.$ So the correct option is A.

5. Verification with stored answer

Stored correct answer: A

Our derived answer: A

They agree.