Answer: D

Given a rod of length \( L = 0.2 \, \text{m} \) with a temperature gradient of \( 416 \, \text{K/m} \) and one end at \( T_1 = 233^\circ \text{C} \):

\(\Delta T = \text{gradient} \times L = 416 \, \text{K/m} \times 0.2 \, \text{m} = 83.2 \, \text{K}\)

The temperature at the other end \( T_2 \) is:

\(T_2 = T_1 + \Delta T = 233^\circ \text{C} + 83.2 \, \text{C} = 316.2^\circ \text{C}\)

Thus, the temperature at the other end is \( 316.2^\circ \text{C}\).