1. Let the radius and height of the right circular cone be $r$ and $h$ respectively.

2. Given slant height $l=3$ m. For a right circular cone, $r^2+h^2=l^2=9$ So, $h=\sqrt{9-r^2}$

3. Volume of the cone is $V=\frac{1}{3}\pi r^2 h$ Substitute $h=\sqrt{9-r^2}$: $V(r)=\frac{1}{3}\pi r^2\sqrt{9-r^2}$

4. To maximize $V$, it is enough to maximize $f(r)=r^2\sqrt{9-r^2}$ Instead, maximize $f(r)^2$ for convenience: $f(r)^2=r^4(9-r^2)$ Let $g(r)=r^4(9-r^2)=9r^4-r^6$

5. Differentiate: $g'(r)=36r^3-6r^5=6r^3(6-r^2)$ Set $g'(r)=0$: $6r^3(6-r^2)=0$ So critical points are $r=0 \quad \text{or} \quad r^2=6$ For maximum volume, take $r=\sqrt{6}$

6. Then $h=\sqrt{9-6}=\sqrt{3}$

7. Maximum volume: $V_{\max}=\frac{1}{3}\pi (6)(\sqrt{3})=2\sqrt{3}\pi$

8. Compare with options:

• A: $2\sqrt3\pi$ ✔
• B: $3\sqrt3\pi$ ✘
• C: $6\pi$ ✘
• D: $\frac{4}{3}\pi$ ✘

Therefore, the correct option is A.