The correct answer is A. 264 MN.
The safe load for the column is calculated as follows:
$P_s = \frac{0.85f_c’A_c + f_yAs}{1 + \frac{f_y}{E_s}} = \frac{0.85 \times 5 \times 0.002 \times 0.04 + 250 \times 0.12}{1 + \frac{250}{200000}} = 264 \text{ MN}$
where:
- $P_s$ is the safe load for the column
- $f_c’$ is the permissible stress in concrete
- $A_c$ is the cross-sectional area of the concrete
- $f_y$ is the yield stress of the steel
- $As$ is the cross-sectional area of the steel
- $E_s$ is the Young’s modulus of the steel
Option B is incorrect because it uses the wrong value for $f_c’$. Option C is incorrect because it uses the wrong value for $f_y$. Option D is incorrect because it uses the wrong value for $E_s$.