Start with the shape of soil the plough disturbs. Its cross-section is a trapezium with parallel sides 16 cm and 4 cm and height (depth) 15 cm, so its area is the average of the two parallel sides times the depth.
\[A = \frac{16+4}{2}\times 15 = 150\ \text{cm}^2\]
The soil resists cutting with a force per unit area of 0.71 kg/cm\(^2\), acting in the horizontal direction of travel. The total horizontal resisting force is:
\[F_h = 0.71 \times 150 = 106.5\ \text{kgf}\]
The bullocks, however, pull the plough along a line tilted 45° above the horizontal, not straight along the ground. Only the horizontal part of their pull actually overcomes soil resistance, so if \(P\) is the pull along that tilted line, its horizontal part is \(P\cos45^\circ\), and this must equal \(F_h\).
\[P = \frac{F_h}{\cos45^\circ} = \frac{106.5}{0.707}\]
\[\boxed{P \approx 150\ \text{kgf}}\]