Answer
$3.9\times10^2counts/s$
Work Step by Step
The counting rate is $85\%$ of $35\%$ of the beta particles emitted per second (the activity).
$$(0.035\times10^{-6}Ci)(\frac{3.7\times10^{10}decays/s}{1\;Ci})(\frac{1\;\beta}{decay})(0.35)(0.85)$$
$$=385.3\;counts/s\approx 3.9\times10^2counts/s$$