Answer
$\approx 7920\;\rm year$
Work Step by Step
Since we know that
$$\dfrac{\rm ^{14}C}{\rm ^{12}C}=1.3\times 10^{-12}$$
which means that $\rm ^{14}C$ is so small.
Now we need to find the number of the $\rm ^{12}C$ carbon atoms in this 73 g.
Thus,
$$N_{\rm ^{12}C}=\dfrac{73}{12}\times N_A=\dfrac{73}{12}\times 6.02\times 10^{23}$$
$$N_{\rm ^{12}C}=\bf 3.66\times10^{24}\;\rm atom$$
And hence, the number of the $\rm ^{14}C$ carbon atoms is given by
$$ \rm ^{14}C =1.3\times 10^{-12}\;\;{\rm ^{12}C}$$
Plugging from above;
$$ \rm ^{14}C =1.3\times 10^{-12}(3.66\times10^{24})$$
$$N_{\rm ^{14}C}=\bf 4.76\times10^{12}\;\rm nuclei$$
We know that the activity is given by
$$R=\left|\dfrac{\Delta N}{\Delta t}\right|_{\rm today}=\left|\dfrac{\Delta N}{\Delta t}\right|_0e^{-\lambda t}=7\;\rm decays /second$$
And hence,
$$ 7=\left|\lambda N_{\rm ^{14}C}\right|_0e^{-\lambda t}$$
$$ \dfrac{7}{\left|\lambda N_{\rm ^{14}C}\right|_0}=e^{-\lambda t}$$
Thus,
$$\ln\left[ \dfrac{7}{\left|\lambda N_{\rm ^{14}C}\right|_0}\right] =-\lambda t$$
Solving for $t$;
$$t=\dfrac{-1}{\lambda}\;\ln\left[ \dfrac{7}{\left(\lambda N_{\rm ^{14}C}\right)_0}\right]$$
where $\dfrac{-1}{\lambda}=\tau=\dfrac{T_{\frac{1}{2}}}{\ln 2}$;
$$t=\dfrac{-T_{\frac{1}{2}}}{\ln 2}\;\ln\left[ \dfrac{7}{\left(\dfrac{\ln 2} { T_{\frac{1}{2}}}N_{\rm ^{14}C}\right)_0}\right]$$
Plugging the known;
$$t=\dfrac{-5730\times 365.25\times 86400}{\ln 2}\;\ln\left[ \dfrac{7}{\dfrac{\ln 2} {5730\times 365.25\times 86400} \times4.76\times10^{12}} \right]$$
$$t=\color{red}{\bf 2.499\times 10^{11}}\;\rm s\approx \color{red}{\bf 7920}\;\rm year$$