The forward characteristic of a silicon diode is nearly a straight line above 0.7 V…

Step 1. The dynamic resistance. $r_d = \dfrac{0.88\ \text{V} - 0.78\ \text{V}}{0.060\ \text{A} - 0.020\ \text{A}}$ $= \dfrac{0.10\ \text{V}}{0.040\ \text{A}} = 2.5\ \Omega$.

Step 2. The knee voltage. Going back from the first point to $I = 0$ along the line: $V_{knee}$ $= 0.78\ \text{V} - 0.020\ \text{A} \times 2.5\ \Omega$ $= 0.73$ V.

Step 3. The diode at 40 mA. $V_d$ $= 0.73\ \text{V} + 0.040\ \text{A} \times 2.5\ \Omega$ $= 0.83$ V.

Step 4. The resistor. It takes $5.0 - 0.83 = 4.17$ V at $40$ mA. $R = \dfrac{4.17\ \text{V}}{0.040\ \text{A}} = 104\ \Omega$.

Why a resistor. Above the knee the diode has only a few ohm, so without a series resistor the current would rise until the diode is damaged.