The distance (d) between two points P(x_1,y_1) and Q(x_2,y_2) on the xy plane is given by:
d = sqrt((x_2-x_1)^2 +(y_2-y_1)^2)
In this example: d=61, P= (10,y), Q=(70,14)
Hence: 61 = sqrt((70-10)^2+(14-y)^2)
61^2 = 60^2 + (14-y)^2
(14-y)^2 = 61^2-60^2 = (61+60)(61-60) = 121
y^2-28y+196 = 121
y^2-28y+75=0
(y-25)(y-3)=0
:. y = 25 or 3
NB: Since the question called for a single value of y I suspect there was a typo. For example, if the distance between P and Q was 60 rather than 61, this would lead to a single value for y of 14, with the same logic as above as follows:
60 = sqrt((70-10)^2+(14-y)^2)
60^2 = 60^2 +(14-y)^2
(14-y)^2=0 -> y=14
In this case PQ is a horizontal line at y=14
