$A_s = 3 \times \frac{1}{4}\pi(32^2) = 768\pi ~ \text{mm}^2$
$nA_s = 9(768\pi) = 6912\pi ~ \text{mm}^2$
Assume NA is at the bottom of the flange
$Q_{\text{above NA}} = 600(80)(40) = 1\,920\,000 ~ \text{mm}^3$
$Q_{\text{below NA}} = nA_s(500 - 80) = 9\,201\,169 ~ \text{mm}^3$
Qabove NA < Qbelow NA, therefore, NA is within the web.
$Q_{\text{above NA}} = Q_{\text{below NA}}$
$600x(\frac{1}{2}x) - 300(x - 80)[ \, \frac{1}{2}(x - 80) \, ] = nA_s(500 - x)$
$300x^2 - 150(x - 80)^2 = 6912\pi(500 - x)$
$x = 167 ~ \text{mm}$
$I_{NA} = \dfrac{600x^3}{3} - \dfrac{300(x - 80)^3}{3} + nA_s(500 - x)^2$
$I_{NA} = \dfrac{600(167^3)}{3} - \dfrac{300(167 - 80)^3}{3} + 6912\pi(500 - 167)^2$
$I_{NA} = 3\,273\,562\,384 ~ \text{mm}^4$
$\dfrac{f_s}{n} = \dfrac{M(d - x)}{I_{NA}}$
$\dfrac{f_s}{9} = \dfrac{100(500 - 167)(1000^2)}{3\,273\,562\,384}$
$f_s = 91.55 ~ \text{MPa}$
$T = f_s \, A_s = 91.55(768\pi)$
$T = 220.89 ~ \text{kN}$
$C = T$
$C = 220.89 ~ \text{kN}$ answer
Another Solution for finding C
$f_c = \dfrac{Mx}{I_{NA}} = \dfrac{100(167)(1000^2)}{3\,273\,562\,384}$
$f_c = 5.10 ~ \text{MPa}$
$f_{c1} = \dfrac{M(x - 80)}{I_{NA}} = \dfrac{100(167 - 80)(1000^2)}{3\,273\,562\,384}$
$f_{c1} = 2.66 ~ \text{MPa}$
$C = \frac{1}{2}(f_c + f_{c1})t_f \, b_f + \frac{1}{2}f_{c1}(x - t_f)b_w$
$C = \frac{1}{2}(5.10 + 2.66)(80)(600) + \frac{1}{2}(2.66)(167 - 80)(300)$
$C = 220.95 ~ \text{kN}$ (okay)
