Find the two-digit number.

A. Sum of the squares of the two digits of the two digits number is 26.
B. The ratio between the two-digit number and the sum of the digits of that number is 5:2.
C. The digit in ten’s place is 4 less than the digit in unit place.

To determine which statements are sufficient to find the two-digit number, let the two-digit number be represented as $10x+y$, where $x$ is the tens digit and $y$ is the units digit (with $x\in \{1,2,\dots ,9\}$ and $y\in \{0,1,\dots ,9\}$).

Analyzing Statement A:
The sum of the squares of the two digits is 26.
$x^2+y^2=26$
The only pairs of single-digit perfect squares that sum to 26 are 1 and 25 (since $1+25=26$).
This means $\{x,y\}=\{1,5\}$.
So, the number could be either 15 or 51. Since we do not have a unique solution, Statement A alone is not sufficient.

Analyzing Statement B:
The ratio between the two-digit number and the sum of its digits is 5:2.
$\frac{10x+y}{x+y}=\frac{5}{2}$
Cross-multiplying gives:
$2(10x+y)=5(x+y)$
$20x+2y=5x+5y$
$15x=3y$
$5x=y$
Since $x$ must be a non-zero single digit, the only possible value is $x=1$ (which gives $y=5$). Any larger value of $x$ would result in a two-digit value for $y$. Therefore, the number is uniquely 15.
Thus, Statement B alone is sufficient.

Analyzing Statement C:
The digit in the tens place is 4 less than the digit in the units place.
$x=y-4$
This gives several possibilities: 15, 26, 37, 48, and 59. Thus, Statement C alone is not sufficient.

Combining Statements A and C:
From Statement A, the number is either 15 or 51.
From Statement C, the tens digit must be 4 less than the units digit ($y-x=4$).
Checking the two options:
- For 15: $5-1=4$ (Valid)
- For 51: $1-5=-4$ (Invalid)
Therefore, combining A and C uniquely determines the number to be 15.
Thus, Statements A and C together are sufficient.

Conclusion:
The two-digit number can be found using either Statements A and C together, or Statement B alone.