Given the problem, we need to find the correct system of equations that represents the situation of buying two types of tickets by $38$ students with a total cost of $952$ yuan. Let's denote $x$ as the number of type A tickets and $y$ as the number of type B tickets.

Step 1: Total number of tickets bought by students.
Since $38$ students bought tickets, and assuming each student bought exactly one ticket, we have the equation:
$$x + y = 38$$

Step 2: Total cost of the tickets.
Given that type A tickets cost $26$ yuan each and type B tickets cost $20$ yuan each, and the total amount spent was $952$ yuan, we can write the equation for the total cost as:
$$26x + 20y = 952$$

Therefore, combining Step 1 and Step 2, we get the system of equations:
$$\left\{\begin{array}{l} x + y = 38 \\ 26x + 20y = 952 \end{array}\right.$$

This matches with option C, which is the correct system of equations based on the given problem.

Hence, the answer is: $\boxed{C}$.