Answer
$2.722\times 10^{-14}\;\rm kg $
Work Step by Step
We know that the number of nuclei of $\rm ^{35}S$ is given by
$$R=\lambda N$$
where $\lambda=\dfrac{\ln2}{T_{ {\frac{1}{2}}}}$;
Solving for $N_{^{35}S}$;
$$N_{ ^{35}S}=\dfrac{R}{\lambda}=\dfrac{RT_{ {\frac{1}{2}}}}{\ln 2}\tag 1$$
We know that the mass amount of $\rm ^{35}S$ is given by
$$m_{^{35}S}=N_{^{35}S}M_{^{35}S}$$
where $M_{^{35}S}$ is its atomic mass.
Plugging from (1)
$$m_{^{35}S}=\dfrac{RT_{ {\frac{1}{2}}}}{\ln 2}M_{^{35}S}$$
Plugging the known;
$$m_{^{35}S}=\dfrac{(4.28\times 10^4)(87.37\times 86400)}{\ln 2}\times 34.969032 \times 1.67\times 10^{-27}$$
$$m_{^{35}S}=\color{red}{\bf 2.722\times 10^{-14}}\;\rm kg $$
