The work done by a force is given by the integral of the force over the displacement. The force is given by:

$$ F(x) = \alpha + \beta x^2 $$

To find the work done when the object is displaced from $x = 0$ to $x = 1$, we compute:

$$ W = \int_{0}^{1} (\alpha + \beta x^2) \, dx $$

Substituting $\alpha = 1 \, \text{N}$:

$$ W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx $$

Calculating each integral separately:

$ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 $

$ \int_{0}^{1} \beta x^2 \, dx = \beta \left[ \frac{x^3}{3} \right]_{0}^{1} = \beta \left[\frac{1^3}{3} - \frac{0^3}{3}\right] = \frac{\beta}{3} $

Thus, the total work done is:

$$ W = 1 + \frac{\beta}{3} $$

We are given that the work done is 5 J, so:

$$ 1 + \frac{\beta}{3} = 5 $$

Subtract 1 from both sides:

$$ \frac{\beta}{3} = 4 $$

Multiply both sides by 3 to solve for $\beta$:

$$ \beta = 12 \, \text{N/m}^2 $$

Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$. Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.