Java solution:

int sumaTotal =-63;

    for(int i =0; i  <= 3 ; i++){
        for(int j = 0; j <= 3 ; j++){

        int sumaActual = arr.get(i).get(j) + arr.get(i).get(j + 1) + arr.get(i).get(j + 2) + 
        arr.get(i+1).get(j +1) +
        arr.get(i + 2).get(j) + arr.get(i+2).get(j+1) + arr.get(i +2).get(j + 2);
        if(sumaActual > sumaTotal){
            sumaTotal = sumaActual;
        }
        }
    }
    return sumaTotal;
}

c#

public static int hourglassSum(List<List<int>> arr)
{
    int highestSum = int.MinValue;
    for (int x = 0; x <= 3; x++)
    {
        for (int y = 0 ; y <= 3; y++)
        {
            int sum = 0;
            for(int x1 = 0; x1 <= 2; x1++)
            {
                for(int y1 = 0; y1 <= 2; y1++)
                {
                    if (x1 == 1 && y1 == 0) continue;
                    if (x1 == 1 && y1 == 2) continue;
                    sum += arr[x + x1][y + y1];
                }
            }
            highestSum = Math.Max(highestSum, sum);
        }
    }

    return highestSum;
}

}

Here is 2D Array - DS solution in PYthon, java, c++ and c programming - https://programmingoneonone.com/hackerrank-2d-arrays-ds-problem-solution.html

Mi solución en Python 3:

def hourglassSum(arr):
    # Write your code here
    maxi = -1000
    for i in range(0,4):
        for j in range(0,4):
            suma= (arr[i][j]+arr[i][j+1]+arr[i][j+2]+
            arr[i+1][j+1]+
            arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2])
            
            maxi= max(maxi,suma)
    return(maxi)