根据《混规》第 11.1.6 条
$$
\mu_ {\mathrm{N}} = \frac {N}{f _ {\mathrm{c}} A} = \frac {2 0 0 0 \times 1 0 0 0}{1 9 . 1 \times 5 0 0 \times 5 0 0} = 0. 4 2 > 0. 1 5, \gamma_ {\mathrm{RE}} = 0. 8
$$
根据《混规》第6.2.5、6.2.17条
$$
x = \frac {\gamma_ {\mathrm{RE}} \times N}{\alpha_ {1} f _ {\mathrm{c}} b} = \frac {0 . 8 \times 2 0 0 0 \times 1 0 0 0}{1 \times 1 9 . 1 \times 5 0 0} = 1 6 7. 5 \mathrm{mm}, 2 a _ {\mathrm{s}} ^ {\prime} < x < \xi_ {\mathrm{b}} h _ {0}
$$
为大偏心受压
$$
e _ {0} = \frac {M}{N} = \frac {6 0 0 \times 1 0 ^ {6}}{2 0 0 0 \times 1 0 ^ {3}} = 3 0 0 \mathrm{mm}
$$
$$
e _ {\mathrm{a}} = \max \left\{2 0, 5 0 0 / 3 0 \right\} = 2 0 \mathrm{mm}, e _ {\mathrm{i}} = e _ {0} + e _ {\mathrm{a}} = 3 0 0 + 2 0 = 3 2 0 \mathrm{mm}
$$
$$
e = e _ {\mathrm{i}} + h / 2 - a _ {\mathrm{s}} = 3 2 0 + 5 0 0 / 2 - 5 0 = 5 2 0 \mathrm{mm}
$$
$$
\gamma_ {\mathrm{RE}} N e \leq \alpha_ {1} f _ {\mathrm{c}} b x (h _ {0} - \frac {1}{2} x) + f _ {\mathrm{y}} ^ {\prime} A _ {\mathrm{s}} ^ {\prime} (h _ {0} - a _ {\mathrm{s}} ^ {\prime})
$$
$$
0. 8 \times 2 0 0 0 \times 1 0 0 0 \times 5 2 0 \leq 1. 0 \times 1 9. 1 \times 5 0 0 \times 1 6 7. 5 \times (4 5 0 - 0. 5 \times 1 6 7. 5) + 4 3 5 \times A _ {\mathrm{s}} ^ {\prime} \times (4 5 0 - 5 0)
$$
求得 $A_{s}^{\prime} = 1414.6\mathrm{mm}^{2}$
因此选（C）