An incompressible fluid flows steadily through a cylindrical pipe having radius $R$ at point $A$ and $\left(\frac{R}{3}\right)$ at point $B$ further along the direction of flow of liquid. If the velocity at point A is ' V ' then that at point $B$ is

The equation of continuity for an incompressible fluid in steady flow requires that the mass flow rate remains constant, thus $A_A V_A = A_B V_B$.

The cross-sectional area varies with the square of the radius: $A_A = \pi R^2$ and $A_B = \pi \left(\frac{R}{3}\right)^2 = \pi \frac{R^2}{9}$.

Substituting into the continuity equation gives $\pi R^2 V = \pi \frac{R^2}{9} V_B$.

Solving for $V_B$ by canceling $\pi R^2$ yields $V_B = 9V$.

Final answer: $\boxed{9V}$