Differentiate the equation of the curve to find the gradient function.

The equation of the curve is $y = ax^3 - 20x^2 + 10x$.
Differentiate $y$ with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(ax^3) - \frac{d}{dx}(20x^2) + \frac{d}{dx}(10x)$$ $$\frac{dy}{dx} = 3ax^2 - 40x + 10$$

The gradient is given at the point where $x=2$.

$$\frac{dy}{dx}\Big|_{x=2} = 3a(2)^2 - 40(2) + 10$$ $$\frac{dy}{dx}\Big|_{x=2} = 3a(4) - 80 + 10$$ $$\frac{dy}{dx}\Big|_{x=2} = 12a - 70$$

The gradient of the curve at $x=2$ is given as 40.

$$12a - 70 = 40$$

$$12a = 40 + 70$$ $$12a = 110$$ $$a = \frac{110}{12}$$ $$a = \frac{55}{6}$$

The value of $a$ is $\boxed{\frac{55}{6}}$.