Step 1 :

Choose $f_a$ equal to the highest frequency of the input signal. In this case $f_a$ = 1KHz $$f_a = \frac{1}{2πR_FC_1}$$ Let $C_1$ = 0.1μf $$1KHz = \frac{1}{2πR _F×0.1µf}$$ $$R_F = 1.59KΩ$$

Step 2 :

Choose $f_b$ = 10 $f_a$ $$f_b = 10K = \frac{1}{2πR_1C_1}$$ $$10K = \frac{1}{2πR_1×0.1µf}$$ $$R_1 = 159.15Ω$$

Step 3 :

Calculate the values of $C_F$, so that $R_1C_1$ = $R_FC_F$

$$159.15Ω ×0.1µf = 1.59KΩ×C_F$$

$$C_F=0.01µf$$

Step 4 :

Designed circuit diagram