Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product
Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:

$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:

$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$

To solve we take the natural logarithm of both sides:

$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$
$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$

And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:

$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$

which does turn out to be correct.