$A_s = 4 \times \frac{1}{4}\pi (28^2) = 784\pi ~ \text{mm}^2$
$nA_s = 9(784\pi) = 7056\pi ~ \text{mm}^2$
$A_s' = 2 \times \frac{1}{4}\pi (28^2) = 392\pi ~ \text{mm}^2$
$(2n - 1)A_s' = [ \, 2(9) - 1 \, ](392\pi) = 6664\pi ~ \text{mm}^2$
$Q_{\text{above NA}} = Q_{\text{below NA}}$
$300x(\frac{1}{2}x) + (2n - 1)A_s'(x - 64) = nA_s(536 - x)$
$150x^2 + 6664\pi (x - 64) = 7056\pi (536 - x)$
$150x^2 + 13\,720\pi x - 4\,208\,512\pi = 0$
$x = 186.15 \, \text{ and } \, -473.50$
Use $x = 186.15 ~ \text{mm}$
$I_{NA} = \dfrac{300(x^3}{3} + (2n - 1)A_s'(x - 64)^2 + nA_s(536 - x)^2$
$I_{NA} = \dfrac{300(186.15^3)}{3} + 6664\pi(186.15 - 64)^2 + 7056\pi(536 - 186.15)^2$
$I_{NA} = 3\,670\,555\,446 ~ \text{mm}^4$
Bending Stresses
$f_b = \dfrac{Mc}{I}$
Concrete:
$f_c = \dfrac{Mx}{I_{NA}}$
$f_c = \dfrac{150(186.15)(1000^2)}{3\,670\,555\,446}$
$f_c = 7.61 ~ \text{MPa}$ answer
Tension steel:
$\dfrac{f_s}{n} = \dfrac{M(536 - x)}{I_{NA}}$
$\dfrac{f_s}{9} = \dfrac{150(536 - 186.15)(1000^2)}{3\,670\,555\,446}$
$f_s = 128.67 ~ \text{MPa}$ answer
Compression steel:
$\dfrac{f_s'}{2n} = \dfrac{M(x - 64)}{I_{NA}}$
$\dfrac{f_s'}{2(9)} = \dfrac{150(186.15 - 64)(1000^2)}{3\,670\,555\,446}$
$f_s' = 89.85 ~ \text{MPa}$ answer
