A burnt plece of wood found at an archaeological site contained $0.0378\mathrm{g}$ of ${}^{14}\mathrm{C}({\mathrm{C}}_{\frac{1}{2}}=5.70\times {10}^3$ years $)$. If it was determined that the wood was about 17.100 vears old, how much ${}^{14}\mathrm{C}$ was present before it was burned? Be sure your answer has the correct number of significant figures.

Step 1 ：Given the half-life of 14C is $5.70\times {10}^3$ years, we can calculate the decay constant $\lambda$ using the formula $\lambda =\frac{\ln (2)}{T_{1/2}}$.

Step 2 ：Substituting the given half-life into the formula, we get $\lambda =\frac{\ln (2)}{5.70\times {10}^3\text{ years}}=1.22\times {10}^{-4}{\text{ years}}^{-1}$.

Step 3 ：The amount of a radioactive isotope remaining after a certain time can be calculated using the formula $N=N_0\times e^{-\lambda t}$, where $N$ is the final amount of the isotope, $N_0$ is the initial amount of the isotope, $\lambda$ is the decay constant, and $t$ is the time elapsed.

Step 4 ：Substituting the given time (17,100 years), the final amount of 14C (0.0378 g), and the calculated decay constant into the formula, we get $0.0378\text{ g}=N_0\times e^{-1.22\times {10}^{-4}{\text{ years}}^{-1}\times 17,100\text{ years}}$.

Step 5 ：Solving for $N_0$, we get $N_0=\frac{0.0378\text{ g}}{e^{-1.22\times {10}^{-4}{\text{ years}}^{-1}\times 17,100\text{ years}}}=\frac{0.0378\text{ g}}{0.135}=0.280\text{ g}$.

Step 6 ：$\boxed{{\displaystyle N_0=0.280\text{ g}}}$ is the initial amount of 14C.