$A_1 = \frac{1}{2}(80)(40) = 1600 \, \text{ mm}^2$

$x_1 = -\frac{2}{3}(80) = -\frac{160}{3} \, \text{ mm}$

$A_2 = \frac{1}{2}(60b) = 30b \, \text{ mm}$

$x_2 = \frac{2}{3}b \, \text{ mm}$

The centroid will lie on the y-axis, thus,

$\bar{x} = 0$

Hence,

$A\bar{x} = \Sigma ax$

$0 = 1600(-\frac{160}{3}) + 30b(\frac{2}{3}b)$

$\frac{2(30)}{3}b^2 = \dfrac{160(1600)}{3}$

$b = 65.32 \, \text{ mm}$ *answer*