Answer
The magnitude of $A$ is $~~3.6~m$
Work Step by Step
$A = a\hat{i} + b\hat{j}$
$B = c\hat{i} + d\hat{j}$
We can find the sum of the two vectors:
$A+B = (a+c)\hat{i} + (b+d)\hat{j}$
Since $A$ lies on the x-axis, then $b = 0$
Since $A+B$ lies on the y-axis, then $(a+c) = 0$
Then:
$A+B = d~\hat{j}$
The magnitude of $A$ is $\vert a \vert$
The magnitude of $A+B$ is $\vert d \vert$ which is equal to $\vert 2a \vert$
Note that $\vert c \vert = \vert a \vert$
We can find the magnitude of A:
$\sqrt{c^2+d^2} = 8.0~m$
$\sqrt{(a)^2+(2a)^2} = 8.0~m$
$\sqrt{5a^2} = 8.0~m$
$5a^2 = (8.0~m)^2$
$a^2 = \frac{(8.0~m)^2}{5}$
$\vert a \vert = \sqrt{\frac{(8.0~m)^2}{5}}$
$\vert a \vert = 3.6~m$
The magnitude of $A$ is $~~3.6~m$
