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The correct answer is $52$. The mean of a data set is calculated by dividing the sum of the data values by the number of values. It’s given that data set A consists of $10$ values, $9$ of which are shown. Let $x$ represent the $10\text{th}$ data value in data set A, which isn’t shown. The mean of data set A can be found using the expression $\frac{43+45+44+43+38+39+40+46+40+x}{10}$, or $\frac{378+x}{10}$. It’s given that the mean of the $9$ values shown is $42$ and that the mean of all $10$ numbers is greater than $42$. Consequently, the $10\text{th}$ data value, $x$, is larger than $42$. It’s also given that the data values in data set A are positive integers less than $60$. Thus, $42<x<60$. Finally, it’s given that the mean of data set A is an integer. This means that the sum of the $10$ data values, $378+x$, is divisible by $10$. Thus, $378+x$ must have a ones digit of $0$. It follows that $x$ must have a ones digit of $2$. Since $42<x<60$ and $x$ has a ones digit of $2$, the only possible value of $x$ is $52$. Since $52$ is larger than any of the integers shown, the largest integer from data set A is $52$.