To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\mathrm{R}^{-3 / 2}$.

According to Newton's law of universal gravitation, the force of attraction $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $$ F = \frac{G m_1 m_2}{r^2}, $$ where $G$ is the gravitational constant.

However, in this particular case, the force of attraction is given to be proportional to $\mathrm{R}^{-3 / 2}$, so we can write $$ F \propto \frac{1}{R^{3/2}}. $$

Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in orbit must be provided by this gravitational force. Hence, we can write that $$ \frac{m v^2}{R} \propto \frac{1}{R^{3/2}}, $$ where $m$ is the mass of the planet and $v$ is its orbital speed.

Simplifying this, we get $$ v^2 \propto \frac{1}{R^{1/2}}. $$

Now, the speed $v$ can be related to the period T through the circumference of the orbit, which is given by $2\pi R$. The orbital speed is the circumference divided by the period: $$ v = \frac{2\pi R}{T}. $$

Substituting this into our proportionality, we get

$$ \left( \frac{2\pi R}{T} \right)^2 \propto \frac{1}{R^{1/2}}, $$

which simplifies to

$$ \frac{4\pi^2 R^2}{T^2} \propto \frac{1}{R^{1/2}}. $$

Solving for $T^2$, we get

$$ T^2 \propto \frac{R^{2 + 1/2}}{4\pi^2}, $$

so $$ T^2 \propto R^{5/2}. $$

Therefore, the correct option is Option C

$$ T^2 \propto R^{5/2}. $$