But that's too fast!

Let $m_1$ and $m_2$ be the masses of the sun and planet respectively.

First lets find out the center of mass of the planet-sun system which is at a distance of say $r_1$ from the sun and $r_2$ from the planet. The position of the center of mass would be $\dfrac{m_2\left(r_1 + r_2\right)}{\left(m_1 + m_2\right)}$ .

Let the centripetal force on sun be $F_{C_1}$ and on planet be $F_{C_2}$ about the center of mass. Let $F_G$ be the gravitational force between them.

Now $F_{C_1} = F_G = F_{C_2}$ .
So, $m_1 r_1 {\omega}^2 = m_2 r_2 {\omega}^2$ (Angular velocity remains same).
$\implies m_1 r_1 = m_2 r_2$ .

Again, $F_{C_2} = F_G\\ \\ \implies m_2 r_2 {\omega}^2 = \dfrac{G m_1 m_2}{{(r_1 + r_2)}^2}\\ \\ \implies {\omega}^2 = \dfrac{G m_1 m_2 × (r_1 + r_2)}{{(r_1 + r_2)}^2 × m_2 r_2 × (r_1 + r_2)}\\ \\ \implies {\omega}^2 = \dfrac{Gm_1(m_2r_1 + m_2 r_2)}{{(r_1 + r_2)}^3 × m_2 r_2}\\ \\ \implies {\omega}^2 = \dfrac{Gm_1(m_2r_1 + m_1 r_1)}{{(r_1 + r_2)}^3 × m_2 r_2}\\ \\ \implies \omega = \sqrt{\dfrac{G(m_1 + m_2)}{{(r_1 + r_2)}^3}}\\ \\ \implies \dfrac{2\pi}{T} = \sqrt{\dfrac{G(m_1 + m_2)}{{(r_1 + r_2)}^3}}\\ \\ \implies T = \boxed{2\pi\sqrt{\dfrac{{(r_1 + r_2)}^3}{G(m_1 + m_2)}}}$ .

Now we have $m_2 = m_1/6$ (because $V_1 = 3V_2$ and ${\rho}_1 = 2 {\rho}_2$ ). Also $r_1 + r_2 = 90000$ m.

Plugging in the values,
$T = 2 \pi \sqrt{\dfrac{{(90000)}^3 × 6}{2×{10}^{30}×6.67 × {10}^{-11}×7}}\\ \\ =0.013598$ seconds.
= $\color{#3D99F6}{\boxed{13.59852}}$ milliseconds.