Some distant star is to be observed by some telescope of diameter...

The angular resolution $\Delta \theta$ of a telescope is given by the formula:

$\Delta \theta = \dfrac{1.22 \lambda}{a}$

where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.

Given:
$\Delta \theta = 3.0 \times 10^{-7}$ rad
$\lambda = 500 \text{ nm} = 500 \times 10^{-9}$ m $= 5 \times 10^{-7}$ m

Substituting the values into the formula:

$3.0 \times 10^{-7} = \dfrac{1.22 \times 5 \times 10^{-7}}{a}$

$a = \dfrac{1.22 \times 5 \times 10^{-7}}{3.0 \times 10^{-7}}$

$a = \dfrac{6.1}{3.0}$ m

$a = 2.0333...$ m

Converting the diameter into centimeters:

$a = 2.0333... \times 100$ cm $= 203.33$ cm

Rounding to the nearest integer, we get $203$ cm.

Answer: $203$