How do I calculate the normality of NaBrO3? Find the mass of $\ce{NaBrO3}$ required to prepare $\pu{150 ml}$ of $\pu{0.75 N}$ solution based on the reaction
  $$\ce{BrO3- + 6 H+ + 6e- -&gt; Br- + 3 H2O}$$


My analysis:

$\pu{150 ml}$ solution of $\pu{0.75 N}$ means $0.1125$ equivalents of $\ce{NaBrO3}$ are present in the solution. Since, according to the given equation, n-factor (of equivalence) of $\ce{NaBrO3}$ is $6$. So $0.1125$ equivalents means $0.01875$ moles of $\ce{NaBrO3}$. The molecular mass of $\ce{NaBrO3}$ is $151$. Thus, I get the mass as $\pu{2.83 g}$ approximately. Am I correct? 


  Since the book says the answer should be $\pu{1.42 g}$
To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).