Assumptions. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.
Estimating the volume of the tire and cartridge. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is.
$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$
Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$.
If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.
$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$
Estimating the pressure. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure
with a van der Waals calculation to see if nonideality makes any difference.
$$\begin{align}
P_{ideal} &= \frac{n R T}{V}\\
&= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
(298~\mathrm{K})}{1.02~\mathrm{L}}\cdot
\frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
&\approx 129~\mathrm{psi}\\
P_{vdw} &= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\
&= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
(298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol})
(0.04267~\mathrm{L~mol^{-1}})}\\
&~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})}
{(1.02~\mathrm{L})^2}\\
&= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
&\approx 124~\mathrm{psi}\\
\end{align}$$
A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$.
