2.29. The lateral surface area of a regular triangular pyramid is 3 times the area of the base. The area of the circle inscribed in the base is numerically equal to the radius of this circle. Find the volume of the pyramid.
2.29. The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} \cdot S O$ (Fig. 2.24). Let the side of the base be $a$, then

$$
S_{\text{base}}=\frac{a^{2} \sqrt{3}}{4}, O D=\frac{a \sqrt{3}}{6} \text{. By the condition, }
$$

Fig. 2.24
$\pi O D^{2}=O D$, or $O D=\frac{1}{\pi}$, hence $\frac{a \sqrt{3}}{6}=\frac{1}{\pi}$, i.e., $a=\frac{6}{\pi \sqrt{3}}$ and $S_{\text {base }}=\frac{3 \sqrt{3}}{\pi^{2}}$. Considering that $\frac{S_{\text {lateral }}}{S_{\text {base }}}=3$, we get $S_{\text {lateral }}=3 S_{\text {base }}=\frac{9 \sqrt{3}}{\pi^{2}}$. On the other hand, $S_{\text {lateral }}=3 \cdot \frac{1}{2} a \cdot S D=\frac{9 S D}{\pi \sqrt{3}}$ and, therefore, $\frac{9 \sqrt{3}}{\pi^{2}}=\frac{9 S D}{\pi \sqrt{3}}$, i.e., $S D=\frac{3}{\pi}$. From $\triangle S O D$ we find $S O=\sqrt{S D^{2}-O D^{2}}=$ $=\sqrt{\frac{9}{\pi^{2}}-\frac{1}{\pi^{2}}}=\frac{2 \sqrt{2}}{\pi}$. Thus, $V=\frac{1}{3} \cdot \frac{3 \sqrt{3}}{\pi^{2}} \cdot \frac{2 \sqrt{2}}{\pi}=\frac{2 \sqrt{6}}{\pi^{3}}$. Answer: $\frac{2 \sqrt{6}}{\pi^{3}}$.