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, $1 parsec \approx 3.09 \times 10^{16}\; m$