Let A = (x,y) be any point in that rectangular plane. The length of the position vector p is $p = \sqrt{x^{2} + y^{2}}$
Therefore,
$E(p^{2}) = E(x^{2}+y^{2}) = E(x^{2})+E(y^{2})$ ...........(1)
Now, x and y are random variables distributed uniformly over (0,1) and (0,2) respectively.
We know, that for a random variable X distributed uniformly over (a,b),
$E(X) = \frac{a+b}{2}$ and $Var(X) = \frac{(b-a)^{2}}{12}$
Also, for any random variable, in general, $Var(X) = E(X^{2}) - E(X)^{2}$
For the random variable x,
$E(x) = \frac{0+1}{2} = \frac{1}{2}$
$Var(x) = \frac{(1-0)^{2}}{12} = \frac{1}{12}$
$\therefore E(x^{2}) = Var(x) + E(x)^{2} = \frac{1}{12}+\frac{1}{4} = \frac{1}{3}$ ...........(2).
For the random variable y,
$E(y) = \frac{0+2}{2} = 1$
$Var(y) = \frac{(2-0)^{2}}{12} = \frac{1}{3}$
$\therefore E(y^{2}) = Var(x) + E(x)^{2} = \frac{1}{3}+1 = \frac{4}{3}$ ...........(3)
Using (2) and (3) in (1), we get:-
$E(p^{2}) = E(x^{2}) + E(y^{2}) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$
Hence, the answer is D.