Electronic – find good IV curves

Although it can be solved using superposition I think it's much easier to use Thevenin/Norton equivalents:

What you have are 3 Thevenin sources in parallel (i.e. voltage sources that have a resistor connected in series).

A Thevenin source can easily be transformed into an equivalent Norton source (i.e. a current source that has a resistor connected in parallel):
\$R_{Norton}=R_{Thevenin}\$ and \$I_{Norton}=\frac{V_{Thevenin}}{R_{Thevein}}\$

Do this for each of the 3 Thevenin sources and you get 3 Norton sources in parallel, i.e. you get 3 current sources and 3 resistors connected in parallel.

3 current sources in parallel can be replaced by 1 current source whose current value is the sum of the 3 single sources: \$I_{tot}=I_1+I_2+I_3\$

3 resistors in parallel can be replaced by one resistor whose resistance value is also easily computed: \$\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\$.

Yes your answer is correct.

I would however have shown the equations for finding the first two voltages. In a test the method usually gets you more marks than the value.

BTW: There are almost always more than one way to solve these kinds of problems. As such, it is prudent, and a good habit to get into, to do so to verify your work.

In this example. you know

\$R_{TOTAL} = (V_S-V_d)/I_D = (20-2.1)/0.0075 = 2386.67\Omega\$

So

\$R_3 = 2386.67 - 1100 - 450 = 836.67\Omega\$

Addition: Since the assignment also asks you to verify KVL, it may be prudent to work your answer the other way around. Using KVL as part of your initial math method does not "prove" it per se.

So finding R3 from the above, then calculating the voltages across reach resistor to show that \$V_{R1} + V_{R2} + V_{R3} + V_D = V_S\$ would be a better way to answer it.