Let the two integers be $x$ and $y.$
Given that:
- $x+y=26\qquad \to (1)$
- $xy=165\qquad \to(2)$
To be find$:x-y=?$
We know that $(x-y)^{2}=x^{2}+y^{2}-2xy$
$\implies(x-y)^{2}=x^{2}+y^{2}+2xy-4xy$
$\implies(x-y)^{2}=(x+y)^{2}-4xy$
Put the values from equation $(1)$ and $(2)$ and we get
$\implies(x-y)^{2}=(26)^{2}-4\times165$
$\implies(x-y)^{2}=676-660$
$\implies(x-y)^{2}=16$
$\implies(x-y)=\sqrt{16}$
$\implies x-y=\pm4$
$\implies x-y=4$ $(or)$ $x-y=-4$
$-4$ is ruled out as both the sum and product of the numbers are positive meaning both the numbers must be positive.
So, $(C)$ is the correct choice.