d
Diameter of Earth's orbit \(=3 \times 10^{11} m\)

Radius of Earth's orbit, \(r=1.5 \times 10^{11} m\)

Let the distance parallax angle be \(1^{\prime \prime}=4.847 \times 10^{-6}\) rad.

Let the distance of the star be \(D\)

Parsec is defined as the distance at which the average radius of the Earth's orbit subtends an angle of \(1^{\pi}\) \(\therefore\) We have \(\quad \theta=\frac{r}{D}\)

\(D=\frac{r}{\theta}=\frac{1.5 \times 10^{11}}{4.847 \times 10^{-6}}\)

\(=0.309 \times 10^{-6} \approx 3.09 \times 10^{16} m\)

Hence, \(1 parsec \approx 3.09 \times 10^{16}\; m\)