By blackening the surface of the cylinders, their emissivity $\varepsilon$ is increased. When B is blackened, the thermal radiation it emits will increase in accordance with the Stefan-Boltzmann law $P = A \varepsilon \sigma T^4$ (where $A$ is area, $\sigma$ the Stefan-Boltzmann constant, and $T$ temperature). This results in the thermoscope becoming hotter and the measured temperature increasing. When A is also blackened its absorptivity will increase (as this is equal to its emissivity). Consequently, the thermoscope now loses more heat to A and the measured temperature is approximately the same as it was originally.
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