A cube with side length $10$ is divided into two cuboids with integral side lengths by a straight cut. Afterwards, one of these two cuboids is divided into two cuboids with integral side lengths by another straight cut.
What is the smallest possible volume of the largest of the three cuboids?

1. **Initial Setup**: We start with a cube of side length \(10\). The volume of the cube is:
   \[
   V = 10^3 = 1000
   \]

2. **First Cut**: We need to divide the cube into two cuboids with integral side lengths. Let the dimensions of the first cuboid be \(a \times 10 \times 10\) and the second cuboid be \((10-a) \times 10 \times 10\), where \(a\) is an integer between 1 and 9.

   The volumes of these cuboids are:
   \[
   V_1 = a \times 10 \times 10 = 100a
   \]
   \[
   V_2 = (10-a) \times 10 \times 10 = 1000 - 100a
   \]

3. **Second Cut**: We now need to divide one of these cuboids further into two smaller cuboids with integral side lengths. We will consider both cases and find the smallest possible volume of the largest cuboid.

   - **Case 1**: Cut the cuboid with volume \(100a\). Let the dimensions of the resulting cuboids be \(b \times 10 \times 10\) and \((a-b) \times 10 \times 10\), where \(b\) is an integer between 1 and \(a-1\).

     The volumes of these cuboids are:
     \[
     V_{1,1} = b \times 10 \times 10 = 100b
     \]
     \[
     V_{1,2} = (a-b) \times 10 \times 10 = 100a - 100b
     \]

     The largest volume among the three cuboids is:
     \[
     \max(100b, 100a - 100b, 1000 - 100a)
     \]

   - **Case 2**: Cut the cuboid with volume \(1000 - 100a\). Let the dimensions of the resulting cuboids be \(c \times 10 \times 10\) and \((10-a-c) \times 10 \times 10\), where \(c\) is an integer between 1 and \(10-a-1\).

     The volumes of these cuboids are:
     \[
     V_{2,1} = c \times 10 \times 10 = 100c
     \]
     \[
     V_{2,2} = (10-a-c) \times 10 \times 10 = 1000 - 100a - 100c
     \]

     The largest volume among the three cuboids is:
     \[
     \max(100c, 1000 - 100a - 100c, 100a)
     \]

4. **Optimization**: We need to minimize the maximum volume among the three cuboids. We will consider different values of \(a\) and find the corresponding values of \(b\) and \(c\) that minimize the largest volume.

   - For \(a = 3\):
     \[
     V_1 = 300, \quad V_2 = 700
     \]
     - Cut \(V_1\) into \(100b\) and \(300 - 100b\):
       \[
       \max(100b, 300 - 100b, 700)
       \]
       The smallest maximum volume is 700.

   - For \(a = 4\):
     \[
     V_1 = 400, \quad V_2 = 600
     \]
     - Cut \(V_1\) into \(100b\) and \(400 - 100b\):
       \[
       \max(100b, 400 - 100b, 600)
       \]
       The smallest maximum volume is 600.

   - For \(a = 5\):
     \[
     V_1 = 500, \quad V_2 = 500
     \]
     - Cut \(V_1\) into \(100b\) and \(500 - 100b\):
       \[
       \max(100b, 500 - 100b, 500)
       \]
       The smallest maximum volume is 500.

   - For \(a = 6\):
     \[
     V_1 = 600, \quad V_2 = 400
     \]
     - Cut \(V_1\) into \(100b\) and \(600 - 100b\):
       \[
       \max(100b, 600 - 100b, 400)
       \]
       The smallest maximum volume is 600.

   - For \(a = 7\):
     \[
     V_1 = 700, \quad V_2 = 300
     \]
     - Cut \(V_1\) into \(100b\) and \(700 - 100b\):
       \[
       \max(100b, 700 - 100b, 300)
       \]
       The smallest maximum volume is 700.

   - For \(a = 8\):
     \[
     V_1 = 800, \quad V_2 = 200
     \]
     - Cut \(V_1\) into \(100b\) and \(800 - 100b\):
       \[
       \max(100b, 800 - 100b, 200)
       \]
       The smallest maximum volume is 800.

   - For \(a = 9\):
     \[
     V_1 = 900, \quad V_2 = 100
     \]
     - Cut \(V_1\) into \(100b\) and \(900 - 100b\):
       \[
       \max(100b, 900 - 100b, 100)
       \]
       The smallest maximum volume is 900.

From the above calculations, the smallest possible volume of the largest of the three cuboids is \(500\).

The final answer is \(\boxed{500}\)