Calculating Average Rate
You haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:

$$\ce{CaCO3 + 2HCl -&gt; CaCl2 + H2O + CO2}$$

By definition rate of consumption of hydrochloric acid over time $\Delta t$ is:

$$r=\frac{\Delta c(\ce{HCl})}{\Delta t}$$

Since all calcium carbonate reacted completely:

$$\Delta c(\ce{HCl}) = \frac{\Delta n(\ce{HCl})}{V} = \frac{2n(\ce{CaCO3})}{V} = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})}$$

where $m$ is mass, $M$ - molar mass, $V$ - volume.

And the average rate is:

$$r = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})\Delta t} = \frac{2\cdot\pu{3.45 g}}{\pu{1 L}\cdot\pu{100.09 g mol-1}\cdot\pu{4.50 min}\cdot\pu{60 s min-1}} = \pu{2.55e-4 mol L-1 s-1}$$

Also, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.