I have a question where it says " Assume p(x,y) is distributed uniformly in the rectangular area between x in [1-4] and y in [2-4] and 0 elsewhere."
It asks what is the value of the density p(x,y)for (x,y) inside the rectangular region.
We calculate the surface of the rectangle which is 2.3=6 but how come probability density function gives a value greater than 1 ? (6 in this case). What does this outcome mean ?
This specific probability density function does not give a value greater than 1. You first divide by 6, because integrating a probability density function always yields $1$, so $p(x,y)=\frac{1}{6}$:
$$\int_2^4\int_1^4p(x,y)dxdy=\int_2^4\int_1^4\frac{1}{6}dxdy=1$$
In the case where the area would be smaller than $1$ we would have (for a uniform distribution) that $p(x,y)>1$, but that's to conform to the rule that the integral is equal to $1$. The intuition is obtained from the fact that $P=\int_Ap(x,y)dxdy\in [0,1]$ for some area subset $A\subseteq [1,4]\times[2,4]$. If we look for example at $p_2(x,y)=1$ defined on $[1,2]\times[2,3]$, then $A\subseteq [1,2]\times[2,3]\subseteq[1,4]\times[2,4]$ has a higher probability for $p_2(x,y)=1$ than for $p(x,y)=\frac{1}{6}$.