Section Modulus calculation and optimization

Section 1:

Calculation for Section 1:

Iox= 2.025e-006m4 =2025000 mm4 (IoxG value from above image)
Y= 430.17 mm (Measured value fom image 1)
S= I/Y = 2025000 mm4 /430.17 mm = 4707.44 mm3

Yield moment My = S × σy, where σy is the yield strength of the material which is 520 MPa or N/mm2 as mentioned in question.
M= S × σy = 4707.44 mm x 520 N/mm2 = 2447868.8 N mm = 2.4e6 N mm.

To increase the strength of the above section, I increased 3mm the distance between the outer panel geometry to inner panel geometry then repeated the above said procedures then we have got much better strength value. See below section 2 calculation part and images.

Section 2:

Calculation for Section 2:

Iox= 2.556e-006m4=2550000 mm4 (IoxG value from above image)
Y= 430.17 mm (Measured value fom image 1)
S= I/Y = 2550000 mm4 / 430.17 mm = 5927.8 mm3

Yield moment My = S × σy, where σy is the yield strength of the material which is 520 MPa or N/mm2 as mentioned in question.
M= S × σy = 5927.8 mm3 x 520 N/mm2 = 3082456 N mm = 3.08 e6 N mm.

Conclusion:
Thus by increasing the distance between the Inner and outer panel of the Hood assembly we got the higher value of section modulus hence the strength also increased.

Cad model placed in the below path for review and give your valuable feedback.
https://drive.google.com/file/d/1bu2F190rSmYa9OQlvn4Kq_2iAcCv-X-7/view?usp=sharing