(c)

Consider the given diagram,

Assuming area observed and screen both circular, we have

$\theta_1=\theta_2 \Rightarrow \frac{d_1}{f}=\frac{d_2}{h} \Rightarrow \frac{d_2}{d_1}=\frac{h}{f}$

where, $d_1=$ diameter of camera screen and $d_2=$ diameter of area on earth.

Now, $\frac{area \,observed \,on \,earth}{area \,of \,screen}=\frac{A_0}{A}$

$=\frac{\left(\frac{\pi \cdot d_2^2}{4}\right)}{\left(\frac{\pi \cdot d_1^2}{4}\right)}=\frac{d_2^2}{d_1^2}$

$\Rightarrow \quad \frac{A_0}{A}=\left(\frac{h}{f_1}\right)^2=\left(\frac{500 \times 10^{+3}}{50 \times 10^{-2}}\right)^2$

$=\left(10 \times 10^3 \times 10^2\right)^2=\left(10^6\right)^2=10^{12}$