To find the work done by an ideal heat engine (Carnot engine) operating between two temperatures, we first convert the temperatures to the Kelvin scale:

1. $T_{high} = 150 + 273.15 = 423.15$ K .
2. $T_{low} = 25 + 273.15 = 298.15$ K .

The efficiency ($η$) of an ideal engine with no frictional losses is determined by the temperature of the reservoirs: $η = 1 - (T_{low} / T_{high}) = (T_{high} - T_{low}) / T_{high}$.

Substituting the values: $η = (423.15 - 298.15) / 423.15 = 125 / 423.15 ≈ 0.2954$.

The efficiency is also the ratio of work done ($W$) to the heat taken from the higher temperature reservoir ($Q_{high}$): $η = W / Q_{high}$.

Rearranging to find work: $W = Q_{high} \times η$ $W = 500\text{ J} \times 0.2954 ≈ 147.7\text{ J}$.

Thus, the work done by the engine is approximately 147.7 J, matching Option A.