Step-by-step Solution:

1. Analyze the derivative p'(x): The problem states that p(x) is a real polynomial with a local maximum at x=1 and a local minimum at x=3. At these points of local extrema, the first derivative of the function must be zero. Therefore, p'(1) = 0 and p'(3) = 0. This implies that (x-1) and (x-3) are factors of the polynomial p'(x).

2. Determine the form of p'(x) for the least degree polynomial p(x): For p(x) to be of the least degree, p'(x) must be of the least possible degree. Since p'(x) has at least two roots (x=1 and x=3), its minimum degree is 2 (a quadratic). So, we can write p'(x) in the form: $p'(x) = k(x-1)(x-3)$ for some non-zero constant k. If p'(x) is a quadratic, then p(x) will be a cubic polynomial, which is the polynomial of the least degree that can have two distinct local extrema.

3. Find p(x) by integrating p'(x): First, expand the expression for p'(x): $p'(x) = k(x^2 - 4x + 3)$ Now, integrate p'(x) with respect to x to find p(x): $p(x) = \int k(x^2 - 4x + 3) dx$ $p(x) = k \left( \frac{x^3}{3} - \frac{4x^2}{2} + 3x \right) + C$ $p(x) = k \left( \frac{x^3}{3} - 2x^2 + 3x \right) + C$ where C is the constant of integration.

4. Use the given conditions to find the constants k and C: We are given two conditions: p(1) = 6 and p(3) = 2.

   Using p(1) = 6: $6 = k \left( \frac{1^3}{3} - 2(1)^2 + 3(1) \right) + C$ $6 = k \left( \frac{1}{3} - 2 + 3 \right) + C$ $6 = k \left( \frac{4}{3} \right) + C \quad \cdots(1)$

   Using p(3) = 2: $2 = k \left( \frac{3^3}{3} - 2(3)^2 + 3(3) \right) + C$ $2 = k \left( 9 - 18 + 9 \right) + C$ $2 = k(0) + C$ $C = 2$

5. Solve for k: Substitute C = 2 back into equation (1): $6 = \frac{4k}{3} + 2$ $4 = \frac{4k}{3}$ $k = 3$

   (We can also check the second derivative conditions. p''(x) = k(2x-4). p''(1) = -2k and p''(3) = 2k. For a local max at x=1, p''(1) < 0, so -2k < 0 " which means k > 0. For a local min at x=3, p''(3) > 0, so 2k > 0 which means k > 0. Our value k=3 is consistent with these conditions.)

6. Calculate p'(0): Now we have the complete expression for p'(x): $p'(x) = 3(x-1)(x-3)$ To find p'(0), substitute x=0 into this expression: $p'(0) = 3(0-1)(0-3)$ $p'(0) = 3(-1)(-3)$ $p'(0) = 9$

Final Answer:

The value of p'(0) is 9.