Point source $d =100 m$

$R_c=3 mm$

$(A)$ $P =\sigma AeT ^4$

$=5.67 \times 10^{-3} \times 64 \times 10^{-6} \times 1 \times(2500)^4(e=1 \text { black body })$

$=141.75 w$

Option $(A)$ is wrong

$(B)$ Power reaching to the eye

$=\frac{ P }{4 \pi d ^2} \times\left(\pi R _e^2\right)$

$=\frac{141.75}{4 \pi \times(100)^2} \times \pi \times\left(3 \times 10^{-1}\right)^2$

$=3.189375 \times 10^{-8} W$

Option $(B)$ is correct

$(C)$

$\lambda_{ m } T  = b$

$\lambda_{ m }  \times 2500=2.9 \times 10^{-3}$

$\Rightarrow \lambda_{ m }  =1.16 \times 10^{-6}$

$ =1160 mm$

Option $(C)$ is correct

$(D)$ Power received by one eye of observer $=\left(\frac{ hc }{\lambda}\right) \times \dot{ N }$

$\dot{ N }=\text { Number of photons entering into eye per second }$

$\Rightarrow 3.189375 \times 10^{-8}$

$=\frac{6.63 \times 10^{-34} \times 3 \times 10^x}{1740 \times 10^{-9}} \times \dot{N}$

$\Rightarrow \dot{N}=2.79 \times 10^{11}$