A

To calculate the temperature difference between the walls of the tube when steady thermal conditions have been reached, we will use the basic principles of heat transfer.
1. **Determine the energy emitted by radon per second**: Since we have 100 millicuries (mCi) of radon, we can convert this to disintegrations per second using the relation:
\[ 100\, \text{mCi} = 100 \times 3.7 \times 10^{10}\, \text{disintegrations/sec} = 3.7 \times 10^{12}\, \text{disintegrations/sec} \]
2. **Energy emitted**: Each decay releases 5.5 MeV. To convert this to joules:
\[ 5.5\, \text{MeV} = 5.5 \times 1.6 \times 10^{-13}\, J = 8.8 \times 10^{-13}\, J \]
Therefore, the total energy emitted per second is:
\[ E = 3.7 \times 10^{12} \times 8.8 \times 10^{-13} = 3.256 \times 10^{0}\, J \approx 3.26\, J/s \]
3. **Calculating heat transfer through the wall of the tube**: The thermal conductivity \( k \) of glass is given as 0.025 Cal cm$^{-2}$ s$^{-1}$ C$^{-1}$, which is approximately \( 0.025 \times 4.184\, J/cm^2 s C \approx 0.10401 \text{ J/cm}^2 ext{s C} \) (using \( 1 ext{ Cal} = 4.184 ext{ J} \)).
4. **Surface area of the tube**: The internal diameter (d) is 0.2 cm and the length (L) is 5 cm:
\[ A = \pi d L = \pi (0.2\, cm)(5\, cm) = \pi\, cm^2 \approx 0.6283\, cm^2 \]
5. **Heat transfer equation**: Using Fourier's law:
\[ \dot{Q} = k A \frac{\Delta T}{L} \Rightarrow \Delta T = \frac{\dot{Q} L}{k A} \]
Plugging in the values:
\[ \Delta T = \frac{3.26\, J/s \cdot 5\, cm}{0.10401\, J/cm^2\text{s C} \cdot 0.6283\, cm^2} \approx 24.3 C \]
Therefore, the temperature difference between the walls of the tube when steady-state thermal conditions are reached is approximately \(\Delta T \approx 24.3 C\).
Thus, the correct answer is A.