You can't use Ohm's law to calculate the current through a diode.
However, the resistor is in series with the diode and, as you've learned, series connected circuit elements have identical currents.
Thus, you can use Ohm's Law to find the current through the resistor and diode but, to do that, you need the voltage across the resistor.
When you subtract the diode voltage from the source voltage, by KVL, what remains is the voltage across the resistor \$V_R\$.
Now, if you know the desired current, e.g., \$20mA\$, you can calculate the resistance required for that series current.
Thus, by Ohm's Law, the desired resistor value is:
$$R = \dfrac{V_R}{20mA} = \dfrac{V_S - V_D}{20mA} = \dfrac{9V - 1.7V}{20mA}$$
The voltage across your load is 10V (this is assuming that zener is in breakdown).
So ohms law says that the current through the resistor is 5mA.
You're total current from the source is the current through your series resistor (4kohm).
Again, if we assume zener breakdown, then the voltage across that series resistor is 50V.
So Ohms Law says that current through that resistor is 12.5mA.
The remaining current is the current that goes through the zener.
The reason I say "assuming zener breakdown", is because a zener will go into zener breakdown when the reverse current is greater than the zener knee current. You're knee current was not specified. If the current through the zener is less than the knee current, then the zener is in reverse bias, and the calculations have be done over but this time assuming that the zener is simply a reverse biased diode.
You are correct. Your book or whoever told you that the load current was 7.5mA is wrong.
Best Answer
Lets look at this circuit in a slightly clearer topology - below is exactly the same circuit, but rearranged to put all the elements in a way that height is signifying voltage - the higher up in the diagram, the higher the voltage. (It's a nice analogy which helps visualise what is going on).
simulate this circuit – Schematic created using CircuitLab
So, at the top of the diagram is \$20 + 10 = 30 \mathrm{V}\$. At the bottom is \$0 \mathrm{V}\$. So lets see if we can work out the currents \$I_r\$ and \$I_d\$. I should note that the direction I have drawn the current arrows is the direction that conventional current would have to flow (positive to negative).
Well, for \$I_r\$ it is fairly easy. Ohm's law tells us for a resistor (or any device with 'Ohmic' behaviour), that \$V=I\times R\$, so from this we can work out:
$$I_r = \frac{V}{R} = \frac{30}{50} = 0.6 \mathrm{A}$$
Why? Well, at the top of the diagram is \$30\mathrm{V}\$, and at the bottom is \$0\mathrm{V}\$, so there must be that much voltage across the resistor.
Now lets look at the diode. For a diode to conduct the voltage at the Anode must be higher than the voltage at the cathode (*). Now in this diagram we can see that the Anode is at \$20\mathrm{V}\$, and the cathode is at \$30\mathrm{V}\$, so this means the voltage across the diode \$V_ak = 20 - 30 = -10\mathrm{V}\$. So from this we can see that the diode is reverse biased - the voltage at the anode is negative with respect to the cathode. So we know then that \$I_d = 0 \mathrm{A}\$.
The current can't 'split down the parallel branch', because the diode is reverse biased so is blocking the flow of current - it's acting essentially as an open circuit.
If the diode was placed the other way around (not a good idea!), then current could flow down that branch. But the current that flows would not reduce the amount of current that flows through the resistor. Instead it increases the amount of current drawn from the 10V supply (**).
(*) There is some 'reverse leakage current' for a diode, basically the amount of current that it conducts when reverse biased, but this is pretty small so for simplicity we say it is \$0\$.
(**) Assuming an ideal power supply - one which has no limit on how much current you can draw.