Friction pile length, 300 mm, 250 kN

Given: $D = 0.30$ m, unit skin friction $f_s = 60$ kN/m² (ultimate), allowable load $250$ kN, tip carries 20% of the total load.

Assumption: a factor of safety of 2.5 is applied to the allowable load to obtain the ultimate load.

$$\begin{aligned} Q_u &= 2.5 \times 250 = 625\ \text{kN} \\ Q_s &= 0.80 \times 625 = 500\ \text{kN} \quad (\text{shaft share}) \end{aligned}$$ $$\begin{aligned} Q_s &= f_s\,\pi D\,L = 60 \times \pi \times 0.30 \times L = 56.55\,L \\ L &= \frac{500}{56.55} = 8.84\ \text{m} \end{aligned}$$

Answer: length of pile $\approx 8.9$ m, say 9 m. (If no factor of safety is applied, i.e. the shaft carries 200 kN, $L = 3.54$ m.)