Two planets at mean distance d₁ and d₂ from the sun and their frequencies are n₁ and n₂ respectively then

The correct option is: n₂²d₂³ = n₁²d₁³.

Step-by-step derivation:

According to Kepler's Third Law of Planetary Motion (also known as the Law of Periods), the square of the orbital period ($T$) of a planet is directly proportional to the cube of its mean distance ($d$) from the Sun:
$T^2\propto d^3$
This can be written with a constant of proportionality $k$ as:
$T^2=k{d}^3$

The frequency of revolution ($n$) of a planet is defined as the reciprocal of its orbital period ($T$):
$n=\frac{1}{T}\Rightarrow T=\frac{1}{n}$

Now, substituting the expression for $T$ in terms of $n$ into Kepler's relation:
${\left(\frac{1}{n}\right)}^2=k{d}^3$
$\frac{1}{n^2}=k{d}^3$
Multiplying both sides by $n^2$, we get:
$1=k{n}^2{d}^3$
Rearranging the equation:
$n^2{d}^3=\frac{1}{k}=\text{constant}$

Since the product $n^2{d}^3$ is constant for any planet orbiting the Sun, we can write the relationship for two different planets as:
${n_1}^2{d_1}^3={n_2}^2{d_2}^3$
Or, equivalently:
${n_2}^2{d_2}^3={n_1}^2{d_1}^3$