Answer
$\sigma_{FTG} = 3.33\space MPa$
$b = 52.5\space cm$
Work Step by Step
1) Bearing Stress on the Footing, $\sigma_{FTG}$ :
$\sigma_{FTG} = \frac{P}{A_{POST}} = \frac{40 * 10^3\space N}{120 * 100\space m^2}$
$\sigma_{FTG} = \boxed{3.33\space MPa}\space\space\leftarrow\space ANS1$
2)Footing Dimensions:
$\space\space\bullet$Bearing stress on Soil:
$\sigma_{SOIL} =\frac{P}{A_{FTG}} = \frac{40 * 10^3\space N}{b^2\space m^2} = 145 * 10^3 \space Pa$
Solving for the unknown quantity, b :
$b = \sqrt{\frac{40*10^3}{145*10^3}} = 0.525 \space m = \boxed{52.5 cm^2}\space\space\leftarrow ANS2$