1. Principle used

For a balloon of constant volume, the buoyant force equals the weight of displaced air:

$F_b = \rho_{\text{air}} V g$

So the load-carrying capacity is proportional to the density of air, i.e.

$L \propto \rho$

Using the ideal gas relation for air,

$\rho \propto \frac{P}{T}$

Hence,

$\frac{L_2}{L_1} = \frac{\rho_2}{\rho_1} = \frac{P_2/T_2}{P_1/T_1} = \frac{P_2 T_1}{P_1 T_2}$

2. Given data

Initial load carried:

$L_1 = 185\,\text{kg}$

Initial temperature:

$T_1 = 27^\circ C = 300\,K$

Final temperature:

$T_2 = -7^\circ C = 266\,K$

Initial pressure is normal atmospheric pressure:

$P_1 = 76\,\text{cm Hg}$

Final pressure:

$P_2 = 45\,\text{cm Hg}$

3. Apply the proportionality

$L_2 = L_1 \cdot \frac{P_2 T_1}{P_1 T_2}$

Substitute values:

$L_2 = 185 \cdot \frac{45 \times 300}{76 \times 266}$

Now calculate:

$45 \times 300 = 13500$

$76 \times 266 = 20216$

So,

$L_2 = 185 \cdot \frac{13500}{20216}$

$\frac{13500}{20216} \approx 0.6678$

Thus,

$L_2 \approx 185 \times 0.6678 \approx 123.54\,\text{kg}$

4. Match with options

$\boxed{123.54\,\text{kg}}$

So the correct option is:

$\boxed{\text{D}}$

5. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.