The power $P$ dissipated by tidal friction is equal to the rate of decrease of Earth's rotational kinetic energy. We can approximate this using finite differences over a long time interval $\Delta t$:

$$P \approx -\frac{\Delta E_k}{\Delta t}$$

The change in kinetic energy $\Delta E_k$ is due to a small change in the period, $\Delta T$.

$$\Delta E_k = E_{k, \text{final}} - E_{k, \text{initial}} = \frac{c}{(T+\Delta T)^2} - \frac{c}{T^2}$$

Since $\Delta T \ll T$, we can use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$:

$$\frac{1}{(T+\Delta T)^2} = \frac{1}{T^2(1+\Delta T/T)^2} = T^{-2}(1+\Delta T/T)^{-2} \approx T^{-2}(1-2\Delta T/T)$$

Substituting this back into the expression for $\Delta E_k$:

$$\Delta E_k \approx c \left( \frac{1}{T^2}(1 - \frac{2\Delta T}{T}) - \frac{1}{T^2} \right) = -\frac{2c\Delta T}{T^3}$$

Therefore, the power dissipated is:

$$P \approx -\frac{-\frac{2c\Delta T}{T^3}}{\Delta t} = \frac{2c}{T^3} \frac{\Delta T}{\Delta t}$$

The rate of change of the period is $\Delta T = 16$ s over $\Delta t = 10^6$ years. We convert $\Delta t$ to seconds:

$\Delta t = 10^6 \text{ yr} \times 365.25 \text{ day/yr} \times 24 \text{ h/day} \times 3600 \text{ s/h} \approx 3.156 \times 10^{13}$ s.

Substituting the given values:

$$P \approx \frac{2(1.65 \times 10^{39} \text{ J·s²})}{(8.64 \times 10^4 \text{ s})^3} \left( \frac{16 \text{ s}}{3.156 \times 10^{13} \text{ s}} \right)$$ $$P \approx 2.6 \times 10^{12} \text{ W}$$