Let 1 and 2 designate the bottom and top end of the pipeline respectively.
Given data:

Density of oil                                                          ρ= 850 kg/m³
Diameter of pipe at section 1                                  D_1=30 \mathrm{~cm}=0.3 \mathrm{~m}

Diameter of pipe at section 2                                  D_2=15 \mathrm{~cm}=0.15 \mathrm{~m}

Pressure at section 1                                                    p_1=200 \mathrm{kN} / \mathrm{m}^2=200 \times 10^3 \mathrm{~N} / \mathrm{m}^2

Pressure at section 2                                                  p_2=98 \mathrm{kN} / \mathrm{m}^2=98 \times 10^3 \mathrm{~N} / \mathrm{m}^2

Rate of flow                                                                      Q=50 \text { litres } / \mathrm{s}=50 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}=0.05 \mathrm{~m}^3 / \mathrm{s}

Cross-sectional area at section 1 is          A_1=\frac{\pi}{4} D_1^2=\frac{\pi}{4}(0.3)^2=0.0707 \mathrm{~m}^2

Cross-sectional area at section 2 is            A_2=\frac{\pi}{4} D_2^2=\frac{\pi}{4}(0.15)^2=0.0177 \mathrm{~m}^2

Average velocity at section 1 is                V_1=\frac{Q}{A_1}=\frac{0.05 \mathrm{~m}^3 / \mathrm{s}}{0.0707 \mathrm{~m}^2}=0.707 \mathrm{~m} / \mathrm{s}

Average velocity at section 2 is              V_2=\frac{Q}{A_2}=\frac{0.05 \mathrm{~m}^3 / \mathrm{s}}{0.0177 \mathrm{~m}^2}=2.825 \mathrm{~m} / \mathrm{s}

Applying Bernoulli’s equation between sections 1 and 2 along a streamline, one can write

\frac{p_1}{\rho g}+\frac{V_1^2}{2 g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2 g}+z_2

or                              z_2-z_1=\frac{p_1-p_2}{\rho g}+\frac{V_1^2-V_2^2}{2 g}

or                              z_2-z_1=\frac{200 \times 10^3-98 \times 10^3}{850 \times 9.81}+\frac{0.707^2-2.825^2}{2 \times 9.81}

=12.232 — 0.381=11.851m of oil