Step 1 : Finding values of $R_2$ and $R_3$

$$V_{O(P-P)} = \frac{2R_2}{R_3} × V_{SAT}$$ $$6 = \frac{2R_2}{R_3} × V_{SAT} $$

Let $V_{SAT}$ = 13V and $R_2$ = 10KΩ $$6 = \frac{2 × 10K }{R_3} × 13 $$ $$\boxed{R_3 = 43.33KΩ}$$

Step 2 : Finding values of $R_1$ and $C_1$

Frequency formula is given by,

$$f = \frac{R_3}{4R_1C_1R_2}$$ Given f = 5KHz $$5K = \frac{43.33KΩ}{4R_1C_1 × 10KΩ}$$ $$R_1C_1 = 2.166 × 10^-4$$ Let C1 = 0.1µf $$R_1 = \frac{2.166 × 10^-4}{0.1× 10^-6 }$$ $$\boxed{R_1 = 2.166KΩ}$$

Step 3 : Designed circuit diagram