Determining power and resistor values for a zener voltage regulator
Might as well summarize Jim's observations:







simulate this circuit &ndash; Schematic created using CircuitLab

At the lowest input voltage, which is \$+20\;\textrm{V}\$, the output will be assumed at \$+10\;\textrm{V}\$ (the zener is doing its job) and if the worst case load of \$100\;m\textrm{A}\$ is present, then \$R_1\$ must allow for all of the load's \$100\;m\textrm{A}\$ plus the required zener's \$10\;m\textrm{A}\$, for a total of \$110\;m\textrm{A}\$ through \$R_1\$. Since \$R_1\$ cannot be allowed to drop more voltage than \$20\;\textrm{V}-10\;\textrm{V}=10\;\textrm{V}\$, it must be the case that its value is \$R_1=\frac{10\;\textrm{V}}{110\;m\textrm{A}}\approx 90.91\Omega\$.

From there, you can now take the opposite approach and say that there is no load present and ask what happens to the zener in this case, now that the value for \$R_1\$ is known. (We aren't using a standard value, but an exact value now, since the problem doesn't require standard value resistors.) To get the worst case current into the zener, we also now have to assume  that the input voltage is no longer \$+20\;\textrm{V}\$, but now \$+25\;\textrm{V}\$, too. This leads to a current in \$R_1\$ of \$I_{R_1}=\frac{25\;\textrm{V}-10\;\textrm{V}}{90.91\Omega}\approx 165\;m\textrm{A}\$. So, with no load, all of that current must go through the zener. So the power dissipated by the zener in this case is \$P_{D_1}=10\;\textrm{V}\cdot 165\;m\textrm{A}=1.65\;\textrm{W}\$. As a bonus, the resistor, \$R_1\$, will be dissipating \$P_{R_1}=15\;\textrm{V}\cdot 165\;m\textrm{A}=2.475\;\textrm{W}\$.

The only reason I'm posting this is because Jim hasn't.

EDIT: The above design probably calls for a \$5\;\textrm{W}\$ resistor. You can also get \$2.5\;\textrm{W}\$ zeners. (A 1N5347B is a \$10\;\textrm{V}\$ zener specified for \$5\;\textrm{W}\$.) Also, a standard resistor value of \$91\;\Omega\$ would be so close as to not make much difference. So, the design is achievable with real parts.

However, most folks would find another way. It is a lot of unnecessarily wasted power and it is not hard nor expensive to add an emitter follower NPN BJT and track down a slightly different zener voltage (and/or to make other slight adjustments.)

I don't think The Art of Electronics authors are recommending the design. They just wanted you to figure it out for yourself.