Given data

• Distance of Sun from Earth: $D = 1.5 \times 10^{11}\ \text{m}$
• Angular diameter of Sun: $\theta = 2000''$ where $''$ means arcseconds.

Convert angular diameter into radians

We know: $1^\circ = 3600''$ and $1^\circ = \frac{\pi}{180}\ \text{rad}$

Hence, $1'' = \frac{\pi}{180\times 3600}\ \text{rad}$

Therefore, $\theta = 2000 \times \frac{\pi}{180\times 3600}$

$\theta = \frac{2000\pi}{648000}$

$\theta \approx 9.696 \times 10^{-3}\ \text{rad}$

Use small-angle formula

For small angles, $\theta = \frac{d}{D}$ where $d$ is the actual diameter of the Sun.

So, $d = D\theta$

$d = 1.5 \times 10^{11} \times 9.696 \times 10^{-3}$

Calculate

$d \approx 1.4544 \times 10^9\ \text{m}$

Thus, $d \approx 1.45 \times 10^9\ \text{m}$

Match with options

• A: $2.45 \times 10^{10}$ m
• B: $1.45 \times 10^{10}$ m
• C: $1.45 \times 10^9$ m
• D: $0.14 \times 10^9$ m

The correct option is: $\boxed{\text{C}}$