$\Sigma M_O = 0$
$2P = 1.5P_{st} + 3P_{br}$
$2P = 1.5(\sigma_{st} \, A_{st}) + 3(\sigma_{br} \, A_{br})$
$2P = 1.5 \, [ \, \sigma_{st} \, (900) \, ] + 3 \, [ \, \sigma_{br} \, (300) \, ]$
$2P = 1350 \sigma_{st} + 900 \sigma_{br}$
$P = 675 \sigma_{st} + 450 \sigma_{br}$
$\dfrac{\delta_{br}}{3} = \dfrac{\delta_{st}}{1.5}$
$\delta_{br} = 2\delta_{st}$
$\left( \dfrac{\sigma \, L}{E} \right)_{br} = 2 \left( \dfrac{\sigma \, L}{E} \right)_{st}$
$\dfrac{\sigma_{br} (2)}{83} = 2 \left[ \dfrac{\sigma_{st} (1.5)}{200} \right]$
$\sigma_{br} = 0.6225 \sigma_{st}$
When σst = 150 MPa
$\sigma_{br} = 0.6225(150)$
$\sigma_{br} = 93.375 \, \text{MPa} > 70 \, \text{MPa} \, $ (not ok!)
When σbr = 70 MPa
$70 = 0.6225 \sigma_{st}$
$\sigma_{st} = 112.45 \, \text{MPa} ok!)
Use σst = 112.45 MPa and σbr = 70 MPa
$P = 675 \sigma_{st} + 450 \sigma_{br}$
$P = 675(112.45) + 450(70)$
$P = 107\,403.75 \, \text{N}$
$P = 107.4 \, \text{ kN}$ answer