1. Angular momentum of a planet in circular orbit

For a planet of mass $m$ moving in a circular orbit of radius $r$ around the Sun, $L = mvr$ where $v$ is orbital speed.

From gravitation providing centripetal force, $\frac{GMm}{r^2} = \frac{mv^2}{r}$ so, $v = \sqrt{\frac{GM}{r}}$

Hence, $L = mr\sqrt{\frac{GM}{r}} = m\sqrt{GMr}$ So, $L \propto m\sqrt{r}$

2. Apply given data for planets $A$ and $B$

For planet $A$, $L_A = m_1\sqrt{GMr_1} = L$

For planet $B$, $L_B = m_2\sqrt{GMr_2} = 3L$

Taking ratio, $\frac{L_B}{L_A} = \frac{m_2\sqrt{r_2}}{m_1\sqrt{r_1}} = 3$

Thus, $\sqrt{\frac{r_2}{r_1}} = 3\frac{m_1}{m_2}$

Squaring both sides, $\frac{r_2}{r_1} = 9\left(\frac{m_1}{m_2}\right)^2$

3. Use Kepler's third law

For planets orbiting the same Sun, $T^2 \propto r^3$ Therefore, $T \propto r^{3/2}$

So, $\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2}$

Using $\frac{r_2}{r_1} = 9\left(\frac{m_1}{m_2}\right)^2$ we get $\frac{r_1}{r_2} = \frac{1}{9}\left(\frac{m_2}{m_1}\right)^2$

Hence, $\frac{T_A}{T_B} = \left[\frac{1}{9}\left(\frac{m_2}{m_1}\right)^2\right]^{3/2}$

Now, $9^{3/2} = 27$ so, $\frac{T_A}{T_B} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^3$

4. Match with options

This corresponds to: $\boxed{\frac{1}{27}\left(\frac{m_2}{m_1}\right)^3}$ which is Option D.