The area of the figure bounded by the given pair of curves y = \(x^2\)- x + 3 and y = 3 can be found by first finding the points of intersection of the two curves. Setting the two equations equal to each other, we have:

\(x^2 - x + 3 = 3\)

\(x^2 - x = 0\)

\(x(x - 1) = 0\)

This gives us two solutions: x = 0 and x = 1. These are the x-coordinates of the points of intersection. The y-coordinates are both equal to 3, since y = 3 for both points.

The area between the two curves is given by the integral of the difference between the two functions over the interval [0,1]:

\(\int_0^1 (3 - (x^2 - x + 3)) dx\)

\(= \int_0^1 (x - x^2) dx\)

\(= \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1\)

\(= \left(\frac{1}{2} - \frac{1}{3}\right) - \left(0 - 0\right)\)

\(= \frac{1}{6}\)

So, the area of the figure bounded by the given pair of curves is **1/6 units (sq)**.