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To determine the largest intensity of uniform loading that can be applied to the frame without causing the average normal stress or average shear stress to exceed the given limits, we need to calculate the maximum allowable normal stress and shear stress.

1. Maximum Allowable Normal Stress:
The average normal stress is given by the formula:
σ_avg = P/A
where σ_avg is the average normal stress, P is the applied load, and A is the cross-sectional area of member CD.

The cross-sectional area of member CD is given by:
A = (side length)^2
A = (36mm)^2
A = 1296 mm^2

The maximum allowable normal stress is given as 20 MPa:
σ_avg ≤ 20 MPa

Substituting the values, we get:
P/A ≤ 20 MPa
P/1296 mm^2 ≤ 20 MPa

Solving for P, we get:
P ≤ 20 MPa * 1296 mm^2
P ≤ 25920 MPa*mm^2

Therefore, the largest intensity of uniform loading that can be applied to the frame without causing the average normal stress to exceed 20 MPa is 25920 MPa*mm^2.

2. Maximum Allowable Shear Stress:
The average shear stress is given by the formula:
τ_avg = V/A
where τ_avg is the average shear stress, V is the applied shear force, and A is the cross-sectional area of member CD.

The maximum allowable shear stress is given as 24 MPa:
τ_avg ≤ 24 MPa

Substituting the values, we get:
V/A ≤ 24 MPa
V/1296 mm^2 ≤ 24 MPa

Solving for V, we get:
V ≤ 24 MPa * 1296 mm^2
V ≤ 31104 MPa*mm^2

Therefore, the largest intensity of uniform loading that can be applied to the frame without causing the average shear stress to exceed 24 MPa is 31104 MPa*mm^2.

In conclusion, the largest intensity w of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section b-b to exceed 20 MPa and 24 MPa, respectively, is 25920 MPa*mm^2.