Stresses due to the given load, see figure below for reference

$\sigma_1 = \dfrac{P}{\text{Area}} = \dfrac{100,000}{150(300)}$
$\sigma_1 = 20/9 ~ \text{MPa}$

$f_{bx} = \dfrac{6M_x}{hb^2} = \dfrac{6(100,000 \times 30)}{300(150^2)}$

$f_{bx} = 8/3 ~ \text{MPa}$

$f_{by} = \dfrac{6M_y}{bh^2} = \dfrac{6(100,000 \times 70)}{150(300^2)}$

$f_{by} = 28/9 ~ \text{MPa}$

At corner *C*, both *f*_{bx} and *f*_{by} are tensile stresses, hence, corner *C* is carrying the largest tension

$\sigma_C = -\sigma_1 + f_{bx} + f_{by} = -20/9 + 8/3 + 28/9$
$\sigma_C = 32/9 ~ \text{MPa}$

Additional axial compression load *P*_{2} to eliminate tensile stress anywhere over the cross section.

$\sigma_C - \sigma_2 = 0$
$32/9 - \dfrac{P_2 (1000)}{300(150)} = 0$

$P_2 = 160 ~ \text{kN}$ ← *answer*