When do reactions reach steady state?
This type of consecutive reaction does not really come to steady state for very long, as seen in the figure, i.e there is only a small time over which $d[mTC]/dt \approx 0$.  The decay rate of mTC is faster than that of Mo and so mTc appears quickly and decays slowly. (The amount of mTC is multiplied by 10 in the figure is the amount formed is small]. 

The figure is drawn assuming  $k_1 = \ln(2)/65, \, k_2 =\ln(2)/6 $ and $\ce{[Mo]_0} = 1$ and that the other species are zero at $t=0$ The equations are 
$\displaystyle \ce{[Mo]=[Mo]}_0e^{-k_1t}$ and 
$$\displaystyle \ce{[mTc] = [Mo]}_0\frac{k_1}{k_1-k_2}\left( e^{-k_1t}-e^{-k_2t}  \right)$$ The amount of Tc is found as $\ce {[Tc] = [Mo]_0-[Mo]-[mTc]}$

The maximum amount of mTC, which is assumed to be 'steady state'  is found by differentiating. 

The time at maximum is $\displaystyle t_m= \ln(k_2/k_1)/(k_2-k_1)$ which is 22.7 hours so the 'steady state' is established just before this and lasts a little past this time. How much in either way is somewhat subjective unless you use a definition of say $\pm 5$ % of the maximum etc.