A certain radioactive element disintegrates with a decay constant of$7.9\times {10}^{-10}/sec.$. At a given instant of time, if the activity of the sample is equal to $55.3\times {10}^{11}$disintegration/sec, then number of nuclei at that instant of time  

The activity of a radioactive sample is given by the formula: 

$Activity = Decay Cons\tan t \times Number of Nuclei$

Rearranging the formula to solve for the number of nuclei: 

$Number of Nuclei = \frac{Activity}{Decay Cons\tan t}$

Plugging in the given values: 

Activity $=55.3\times {10}^{11}disintegrations/sec$

$Decay Constant=7.9\times {10}^{-10}sec$

Number of Nuclei = $\frac{55.3\times {10}^{11}}{(7.9\times {10}^{-10})}$

Number of Nuclei = $7.0\times {10}^{21}$

Hence the correct answer is $7.0\times {10}^{21}$.