Total change in grade,
A:
$A = g_1 - g_2$
$A = 0.80 - (-0.40)$
$A = 1.20\%$
Number of meter-stations, n:
$n = \dfrac{\text{total change in grade, }A}{\text{change in grade per meter-station}}$
$n = \dfrac{1.20\%}{0.20\%}$
$n = 6 ~ \text{stations}$
Length of vertical curve, L:
$L = \text{number of meter-stations } ~ \times ~ \text{ length of a meter-station}$
$L = 6 \times 20$
$L = 120 ~ \text{m}$ [ C ] answer