Given the polynomial function:

$$f(x) = ax^3 + bx^2 + cx + 41$$

We are provided the following conditions from the problem:

1. $$f(1) = 40$$

2. $$f^{\prime}(1) = 2$$

3. $$f^{\prime \prime}(1) = 4$$

First, calculate $f(1)$:

$$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$$

Simplifying, we get:

$$a + b + c + 41 = 40$$

Therefore:

$$a + b + c = -1$$

Next, calculate the first derivative $f^{\prime}(x)$:

$$f^{\prime}(x) = 3ax^2 + 2bx + c$$

Given $f^{\prime}(1) = 2$:

$$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$$

Simplifying, we get:

$$3a + 2b + c = 2$$

Next, calculate the second derivative $f^{\prime \prime}(x)$:

$$f^{\prime \prime}(x) = 6ax + 2b$$

Given $f^{\prime \prime}(1) = 4$:

$$f^{\prime \prime}(1) = 6a(1) + 2b = 4$$

Simplifying, we get:

$$6a + 2b = 4$$

Dividing the entire equation by 2:

$$3a + b = 2$$

We now have three equations:

1. $$a + b + c = -1$$

2. $$3a + 2b + c = 2$$

3. $$3a + b = 2$$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$$(3a + 2b + c) - (3a + b) = 2 - 2$$

Which simplifies to:

$$b + c = 0$$

So,

$$c = -b$$

Substitute $c = -b$ into the first equation:

$$(a + b - b = -1)$$

Simplifying, we get:

$$a = -1$$

Now substitute $a = -1$ into the third equation:

$$3(-1) + b = 2$$

Which simplifies to:

$$-3 + b = 2$$

Therefore:

$$b = 5$$

Next, since $c = -b$:

$$c = -5$$

Finally, we need to find $a^2 + b^2 + c^2$:

$$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$$

Simplifying, we get:

$$1 + 25 + 25 = 51$$

Therefore, the answer is:

Option B: 51