The effective temperatures of planets can be derived under the assumptions of their energy input coming solely from the Sun. The relations in Essential Astrophysics are derived from the assumptions of energy balance. First, we assume that the planets are in thermal equilibrium so that the energy into the planet equals the energy out of the planet $E_{\mathrm{out}}=E_{\mathrm{in}}$.

We can estimate these two energies. First the output energy is given by the thermal radiation coming off the planet, which is assumed to be a spherical emitter of radius $R_{p}$, radiating at the effective temperature of the planet, $T_{\mathrm{ep}}$:

$$E_{\mathrm{out}} = 4\pi R_{p}^2 \sigma T_{\mathrm{ep}}^4.$$

The energy input from the Sun is set by the flux of radiation from the Sun that strikes the planet $f_\odot$. The tricky part here is that, from far away, all spheres just look like disks of radius $R_p$ so we need to use the cross sectional area $\pi R_{p}^2$. Another wrinkle is that some of the energy gets reflected off the surface and doesn’t get absorbed by the planet. The fraction of energy that is reflected is given by the albedo $A$, so the fraction that is absorbed is $(1-A)$. Thus, the total energy input is:

$$\begin{eqnarray} E_{in} &=& f_{\odot} \cdot \pi R_p^2 \cdot (1-A),\\ & = & \frac{ L_{\odot}}{4\pi D^2} \cdot \pi R_p^2 \cdot (1-A),\\ & = & \frac{ 4\pi R_{\odot}^2 \sigma T_{\odot}^4 }{4\pi D^2} \cdot \pi R_p^2 \cdot (1-A),\\ & = & \frac{R_{\odot}^2 \sigma T_{\odot}^4 }{ D^2} \cdot \pi R_p^2 \cdot (1-A).\\ \end{eqnarray}$$

In the above, we have replaced the flux with the value from the inverse square law, which gives the flux in terms of the solar luminosity $L_{\odot}$ and the distance between the Sun and the planet in question $D$. Now, we can equate the two energies and solve for $T_{\mathrm{ep}}$.

$$\begin{eqnarray} E_{\mathrm{out}} &= & E_{\mathrm{in}} \\ 4\pi R_{p}^2 \sigma T_{\mathrm{ep}}^4 & = & \frac{R_{\odot}^2 \sigma T_{\odot}^4 }{ D^2} \cdot \pi R_p^2 \cdot (1-A), \mbox{ thus}\\ T_{\mathrm{ep}}^4 & = & T_{\odot}^4 \frac{R_{\odot}^2}{4 D^2} (1-A), \\ T_{\mathrm{ep}} & = & T_{\odot} \left(\frac{R_{\odot}}{2 D}\right)^{1/2} (1-A)^{1/4}. \end{eqnarray}$$

If we plug in the values of $D=1\mbox{ AU}$, $A=0.29$ and $T_{\odot}^4$, we arrive at the temperature for Earth: $T_{\mathrm{ep}}=256\mbox{ K}$. This isn’t what’s in the book (that’s for the visual geometric albedo – Thanks for pointing this out Kaleigh). However, this derivation implicitly assumes absorption and re-radiation across all wavelengths and not the visible, so the Bond albedo is what should be used.