Solution:

select a value of $c$; Let $C=0.1 \mathrm{MF}$ calculate value of $R$ from $f_0$ $ f_0=\frac{1}{2 \pi \sqrt{6} R c}\ $ $ \begin{aligned}\ &\therefore R=\frac{1}{2 \pi \sqrt{6} f_0 c} \\ &\therefore R=1082 \Omega \\ &\text { Let } R=1 \mathrm{k} \Omega\ \end{aligned}\ $ To prevent loading of 0,1 -amp by $R C$ network choose $R_1$ much larger than R $ \text { Let } \begin{aligned}\ R_1 & \geq 10 R \\ R_1 &=10 \times 1 \mathrm{k} \Omega \\ R_1 &=10 \mathrm{k} \Omega\ \end{aligned}\ $ The gain of op-amp in the phase shift oscillator is $2 g$ $ \therefore \frac{R_F}{R_1}=2 \mathrm{~g}\ $ $ \begin{aligned}\ \therefore \quad R F &=29 \times R_1 \\ \therefore R F &=29 \times 10 \mathrm{k} \Omega \\ \therefore \quad R F &=290 \mathrm{k} \Omega\ \end{aligned}\ $