The rate of dissipation of the heat by a black body at temperature...

According to Stefan-Boltzmann law, the rate of heat dissipation by a black body is given by $Q = \sigma A T^4$.

For an ordinary body with emissivity $e$ and temperature $T'$, the rate of heat dissipation is $Q' = e \sigma A T'^4$.

Given $e = 0.25$ and $T' = 3T$, substituting these values gives:

$Q' = 0.25 \times \sigma A (3T)^4$

$Q' = 0.25 \times 81 \sigma A T^4$

$Q' = 20.25 Q$

Therefore, the rate of dissipation is nearly $20 Q$.

Answer: $20\ Q$