For concreteness sake, let's suppose we are computing the moment of inertia of the Koch Snowflake $K$ of uniform density and mass 1 shown on the left below. This snowflake is centered at the origin and has diameter 3 - that is, the maximum distance between two points is three. From here, it would be easy to compute the moment of another Koch flake of different size and uniform density, as long as it's centered at the origin.
The moment of inertia of this flake $K$, as I understand it, is simply
$$\iint\limits_{K} (x^2+y^2) \, dA$$
divided by the area of the snowflake.
There are any number of tools that can compute this numerically. Using Mathematica, I came up with 0.813719. (I could provide code, if desired.)
This value can also be computed exactly using the theory of self-similar integration as described in Bob Strichartz paper Evaluating Integrals Using Self-similarity. Applying this, I computed an exact value of $9/11$ or $0.\overline{81}$. The remainder of this answer will describe that process, though it does require a fairly sophisticated knowledge of self-similarity and measure theory.
We will use the fact that the Koch Snowflake is self-similar - composed of 7 copies of itself as shown in the figure above on the right. The scaling factor for the middle piece is $1/\sqrt{3}$ and it's $1/3$ for the others. Let us suppose that the functions in the IFS mapping $K$ onto the individual pieces are $T_0,T_1,\ldots,T_6$, with $T_0$ mapping onto the central piece. The exact functions in the IFS are
\begin{align}
T_0(x,y) &= \frac{1}{\sqrt{3}}R\left(\frac{\pi}{2}\right)\left(\begin{array}{c}{x\\y}\end{array}\right) \\
T_{i}(x,y) &= \frac{1}{3}\left(\begin{array}{c}{x\\y}\end{array}\right) +
\left(\begin{array}{c}{\cos\left(\frac{\pi}{6} + i\frac{\pi}{3}\right)\\
\sin\left(\frac{\pi}{6} + i\frac{\pi}{3}\right)}\end{array}\right),
\end{align}
where $\varphi$ ranges from $\pi/6$ to $11\pi/6$ in steps of $\pi/3$ and $R(\pi/2)$ represents a rotation through the angle $\pi/2$.
As explained by Strichartz, the integral of a function $f$ defined on a self-similar set can be computed with respect to any self-similar measure $\mu$. Specializing our notation somewhat to deal with this specific problem, a measure $\mu$ on $\mathbb R^2$ is called self-similar with respect to a list of weights $p_0,p_1,\ldots,p_6$, if it satisfies
$$\mu(A) = \sum_{i=0}^6 p_i \mu(T_i^{-1}(A)$$
for every $A\subset\mathbb R^2$. As a result, any integral with respect to $\mu$ will satisfy
$$\int f d\mu = \sum_{i=0}^6 p_i \int f \circ T_i \, d\mu.$$
As explained in Strichartz' paper there is a unique measure $\mu$ that satisfies the self-similarity condition for any given list of probabilities $p_0,p_1,\ldots,p_6$. We need to choose the probabilities so that the measure $\mu$ is uniform. Given an IFS with scaling factors $r_0, r_1, \ldots, r_6$, a uniform measure can be constructed by choosing $p_i = r_i^s$, where $s$ satisfies Moran's equation
$$\sum_{i=0}^6 r_i^s = 1.$$
This yields the probability list $p_0=1/3$ and $p_i = 1/9$ for $i=1,\ldots,6$.
Now, since the total mass of $K$ with respect to $\mu$ is 1, we certainly have that the integral of any constant is just that constant, i.e.
$$\int c \, d\mu = c.$$
We can use the self-similarity of the integral to compute the integrals of $f_1(x,y)=x$ and $f_2(x,y)=y$ as follows. Written down as a list of functions that return ordered pairs, the IFS is
\begin{align}
T_0(x,y) &= \left(-\frac{y}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \\
T_1(x,y) &= \left(\frac{x}{3}+\frac{\sqrt{3}}{2},\frac{y}{3}+\frac{1}{2}\right) \\
T_2(x,y) &= \left(\frac{x}{3},\frac{y}{3}+1\right) \\
T_3(x,y) &= \left(\frac{x}{3}-\frac{\sqrt{3}}{2},\frac{y}{3}+\frac{1}{2}\right) \\
T_4(x,y) &= \left(\frac{x}{3}-\frac{\sqrt{3}}{2},\frac{y}{3}-\frac{1}{2}\right) \\
T_5(x,y) &= \left(\frac{x}{3},\frac{y}{3}-1\right) \\
T_6(x,y) &= \left(\frac{x}{3}+\frac{\sqrt{3}}{2},\frac{y}{3}-\frac{1}{2}\right)
\end{align}
Note that $f_1$ returns just the first component and $f_2$ returns the second. Extracting those first components, multiplying them by the terms in the probability list and adding them up, to apply the self-similar integration identity, we get
$$\int x \, d\mu = \frac{2}{9}\int x \, d\mu - \frac{1}{3\sqrt{3}} \int y \, d\mu.$$
Similarly,
$$\int y \, d\mu = \frac{1}{3\sqrt{3}} \int x \, d\mu + \frac{2}{9}\int y \, d\mu.$$
This leads to a pair of equations in the unknowns
$$\int x \, d\mu \: \text{ and } \int y \, d\mu,$$
with a unique solution that both are zero. From the symmetry of those functions and the domain, this is absolutely correct. A similar procedure may be carried out on the second order terms. We find that
\begin{align}
\int x^2 \, d\mu &= \int (1/3 + 2x^2/27 +y^2/9) \, d\mu \\
\int xy \, d\mu &= -\int xy/27 \, d\mu \\
\int y^2 \, d\mu &= \int (1/3 + x^2/9 + 2y^2/27) \, d\mu.
\end{align}
Solving these for the integrals of $x^2$ and $y^2$ and adding, we obtain the desired result.
Best Answer
In each of the figures below, the cross represents the centre of rotation.
Denote the moment of inertia of the region $S_i$ by $I(S_i)$. The moment of inertia of the fidget spinner is $6I(S_1)+3I(S_4)-6I(S_3)+6I(S_2)$.
Using the parallel axis theorem, it can be deduced that $I(S_4)=I(S_8)+4MR^2$ and that $I(S_1)=I(S_7)+MR^2$. It should be obvious that $I(S_6)=I(S_8)/3$.
The center of mass of region $S_5$ is at a distance of $\frac{R}{2\sqrt{2}}$ from the centre of rotation so we can use the parallel axis theorem again to deduce that $I(S_2)=I(S_5)+MR^2((2-\frac{1}{2\sqrt{2}})^2-1/8)$.
Finally, the distance between the centre of mass of $S_6$ and the centre of rotation turns out to be $\frac{3\sqrt3}{2\pi}$. So $I(S_3)=I(S_6)+MR^2((2-\frac{3\sqrt3}{2\pi})^2-\frac{27}{2\pi})$.
Now all we need to do is to find $I(S_8)$, $I(S_5)$, $I(S_7)$, either by calculating them or by looking them up on Wikipedia.
As it turns out, $I(S_5)=\frac{5}{12}MR^2$, $I(S_7)=MR^2/4$ and $I(S_8)=MR^2/2$.