Draw Cartesian axes labeled $A$ and $B$. You're asking about the fraction of the square $[0,10]\times[0,10]$ that is greater than $4$ units away (measured horizontally or vertically) from the diagonal $A=B$. Geometrically this region consists of two isosceles right triangles with side length $6$, so its total area is $36$. The desired probability is $0.36$.
Byron has already answered your question, but I will attempt to provide a detailed solution...
Let $X$ be a random variable uniformly distributed over $[0,L]$, i.e., the probability density function of $X$ is the following
$$f_X (x) = \begin{cases} \frac{1}{L} & \textrm{if} \quad{} x \in [0,L]\\ 0 & \textrm{otherwise}\end{cases}$$
Let us randomly pick two points in $[0,L]$ independently. Let us denote those by $X_1$ and $X_2$, which are random variables distributed according to $f_X$. The distance between the two points is a new random variable
$$Y = |X_1 - X_2|$$
Hence, we would like to find the expected value $\mathbb{E}(Y) = \mathbb{E}( |X_1 - X_2| )$. Let us introduce function $g$
$$g (x_1,x_2) = |x_1 - x_2| = \begin{cases} x_1 - x_2 & \textrm{if} \quad{} x_1 \geq x_2\\ x_2 - x_1 & \textrm{if} \quad{} x_2 \geq x_1\end{cases}$$
Since the two points are picked independently, the joint probability density function is the product of the pdf's of $X_1$ and $X_2$, i.e., $f_{X_1 X_2} (x_1, x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = 1 / L^2$ in $[0,L] \times [0,L]$. Therefore, the expected value $\mathbb{E}(Y) = \mathbb{E}(g(X_1,X_2))$ is given by
$$\begin{align} \mathbb{E}(Y) &= \displaystyle\int_{0}^L\int_{0}^L g(x_1,x_2) \, f_{X_1 X_2} (x_1, x_2) \,d x_1 \, d x_2\\[6pt]
&= \frac{1}{L^2} \int_0^L\int_0^L |x_1 - x_2| \,d x_1 \, d x_2\\[6pt]
&= \frac{1}{L^2} \int_0^L\int_0^{x_1} (x_1 - x_2) \,d x_2 \, d x_1 + \frac{1}{L^2} \int_0^L\int_{x_1}^L (x_2 - x_1) \,d x_2 \, d x_1\\[6pt]
&= \frac{L^3}{6 L^2} + \frac{L^3}{6 L^2} = \frac{L}{3}\end{align}$$
Best Answer
We have the density function for uniform $x\in[0,L]$ and uniform $y\in[0,x]$ $$ f(x,y)=\frac 1L\cdot\frac 1x $$ so it is $$ \int_0^L\int_0^x (x-y)\frac 1L\cdot\frac 1x\ dy\ dx=\frac L4 $$ which can be confirmed by this Wolfram|Alpha-computation.