With an efficiency of 20%, the bulb will emit 5W of light. The energy of a single 500nm photon is:
\[E = \frac{h \cdot c}{\lambda} = \frac{6.626 \cdot 10^{-34}\text{Js} \cdot 300\text{Mm/s}}{500\text{nm}}=3.97 \cdot 10^{-19} \text{J/photon}\]
Therefore to output 5W=5J/s, we must emit \[ {5\text{J/s} \over 3.97 \cdot 10^{-19} \text{J/photon}} = 1.26 \cdot 10^{19} \text{photons/s} \]
The bulb's power is \[ (2.4\text{V})\cdot(0.7\text{A})=1.68\text{W}\]
Because the bulb is ideal, it emits all that power as radiant flux over its surface.
Because the bulb emits evenly over a sphere-shape, it emits evenly through a \(4\pi \text{sr}\) solid angle. This gives a radiant intensity of \[I = \frac{1.68\text{W}}{4\pi\text{sr}} = 0.133\text{W/sr}\]
The radiant intensity is equal to \[M = \frac{I}{R^2}=\frac{0.133\text{W/sr}}{(5\text{mm})^2} = 5350\text{W/m}^2\]
In 5 minutes, the bulb will emit \[ 1.68\text{W} \cdot 5\text{minutes}\cdot60\text{s/minute}=504\text{J}\]
The irradiance is (assuming eye facing toward light bulb): \[ E = \frac{\Phi}{4\pi} \cdot \frac{\cos{\theta}}{r^2} = \frac{1.68\text{W}}{4\pi\cdot(1\text{m})^2}=0.133\text{W/m}^2\]
At the tale, the irradiance is \[ \frac{(20\%) \cdot (200\text{W})}{4\pi (2\text{m})^2} = 0.796\text{W/m}^2\]
This is equivalent to an illuminance of \(685 \cdot 0.1 \cdot 0.796\text{W/m}^2 = 54.5\text{lux}\)
If both sides are equally illuminated, then the flux received on the screen is equal from both sources. That is \[ \frac{I_s}{r_s^2} = \frac{I_x}{r_x^2} \]
By simple algebra: \[ I_x = \frac{r_x^2}{r_s^2} I_s = \frac{(65\text{cm})^2}{(35\text{cm})^2} \cdot 40\text{cd} = 138\text{cd} \]
\[ I_x = 138\text{cd} \]
Because the light source is diffuse, it emits in a cosine-weighted hemisphere. The integral across this cosine-weighted hemisphere is a solid angle of \(\pi\). Therefore, the radiosity is \[ \pi \text{ sr} \cdot 5000\text{W/sr m}^2 = 15700 \text{W/m}^2 \]
In a second, this emits the energy of \[ (1\text{s}) \cdot (10\text{cm})^2 \cdot ( 15700 \text{W/m}^2) = 157\text{J} \]
The radiant exitance \[ \int_{2\pi} 6000 \text{W/s m}^2 \cdot \cos{\theta} \cdot d\omega = 6000\pi \text{W/m}^2 = 18.8 \text{kW/m}^2\]
This emits a power of \[ (0.1\text{m})^2 \cdot (18.8 \text{kW/m}^2) = 188\text{W} \]
Radiance describes the light source and is the same regardless of observer so yes, the radiance from the sun is the same on Mars as on Earth, assuming they have the same angle to the sun. However, Earth may receive a higher flux than Mars leading to more energy on the surface of the Earth. In actuality, because Mars is closer to the sun than Earth, the increases flux at the surface of the Earth is due to light from the sun and radiated infrared photosn from the earth that reflect back to the surface due to greenhouse gasses in the Earth's atmosphere, which is much thicker than Mars'.