$I = \dfrac{bd^3}{12} = \dfrac{80(160^3)}{12} = 27.31 \times 10^6 \, \text{mm}^4$
$f_v = \dfrac{VQ}{Ib} = \dfrac{40(1000)Q}{(27.31 \times 10^6)(80)}$
$f_v = (18.31 \times 10^{-6})Q$
Layer 1 (20 mm from top and bottom layers)
$Q_1 = 80(20)(70)$
$Q_1 = 112\,000 \, \text{mm}^3$
$f_{v1} = (18.31 \times 10^{-6})(112\,000)$
$f_{v1} = 2.0508 \, \text{ MPa}$ answer
Layer 2 (40 mm from top and bottom layers)
$Q_2 = Q_1 + 80(20)(50)$
$Q_2 = 112\,000 + 80\,000$
$Q_2 = 192\,000 \, \text{mm}^3$
$f_{v2} = (18.31 \times 10^{-6})(192\,000)$
$f_{v2} = 3.5156 \, \text{ MPa}$ answer
Layer 3 (60 from top and bottom layers)
$Q_3 = Q_2 + 80(20)(30)$
$Q_3 = 192\,000 + 48\,000$
$Q_3 = 240\,000 \, \text{mm}^3$
$f_{v3} = (18.31 \times 10^{-6})(240\,000)$
$f_{v3} = 4.3945 \, \text{ MPa}$ answer
Layer 4 (The Neutral Axis, NA)
$Q_4 = Q3 + 80(20)(10)$
$Q_4 = 240\,000 + 16\,000$
$Q_4 = 256\,000 \, \text{mm}^3$
$f_{v4} = ( \, f_v \, )_{max} = (18.31 \times 10^{-6})(256\,000)$
$f_{v4} = ( \, f_v \, )_{max} = 4.6875 \, \text{ MPa}$ answer
Checking:
For rectangular section
$( \, f_v \, )_{max} = \dfrac{3V}{2bd} = \dfrac{3(40)(1000)}{2(80)(160)}$
$( \, f_v \, )_{max} = 4.6875 \, \text{MPa}$ (okay!)
