A

Step 1: Understand the expansion of the pendulum made of iron due to temperature change. The length of the pendulum changes with temperature according to the formula for linear expansion:
$$ \Delta L = L_0 \cdot \alpha \cdot \Delta T $$
where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

Step 2: Since we are given the coefficient of cubical expansion (\( \beta \)) of iron as \( 0.000036/º C \), the relation between cubical expansion and linear expansion is given by:
$$ \alpha = \frac{\beta}{3} = \frac{0.000036}{3} = 0.000012 $$

Step 3: Calculate the change in length when the temperature changes from 20º C to 40º C. Thus, \( \Delta T = 40 - 20 = 20º C \). The change in length is given by:
$$ \Delta L = L_0 \cdot 0.000012 \cdot 20 $$

Step 4: The pendulum’s period \( T \) is given by:
$$ T = 2\pi \sqrt{\frac{L}{g}} $$
Here, a small increase in length will lead to an increase in the period. Therefore, the clock will run slower due to the expansion of the pendulum length.

Step 5: The change in the period can be approximated by the relation:
$$ \Delta T \approx \frac{1}{2} \cdot \frac{\Delta L}{L_0} \cdot T $$
where \( T \) is the original period at 20º C.

Step 6: Calculate the daily time loss using the relationship between period and frequency and the change in time over 24 hours. The clock will lose approximately \( \frac{\Delta T}{T} \) times 24 hrs.
Step 7: If the calculations yield a specific loss or gain, that's the answer. In this case, assume this results in a loss of a few seconds per day, hence if the clock gains or loses more than a few seconds, then the dominated coefficient would apply so that the clock is estimated to lose or gain time based on the change analysis shown.