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The correct answer is $-34$. A system of two linear equations in two variables, $x$ and $y$, has no solution if the lines represented by the equations in the xy-plane are distinct and parallel. Two lines represented by equations in standard form $Ax+By=C$, where $A$, $B$, and $C$ are constants, are parallel if the coefficients for $x$ and $y$ in one equation are proportional to the corresponding coefficients in the other equation. The first equation in the given system can be written in standard form by subtracting $30y$ from both sides of the equation to yield $48x-102y=24$. The second equation in the given system can be written in standard form by adding $16x$ to both sides of the equation to yield $16x+ry=\frac{1}{6}$.  The coefficient of $x$ in this second equation, $16$, is $\frac{1}{3}$ times the coefficient of $x$ in the first equation, $48$. For the lines to be parallel the coefficient of $y$ in the second equation, $r$, must also be $\frac{1}{3}$ times the coefficient of $y$ in the first equation, $-102$. Thus, $r=\frac{1}{3}(-102)$, or $r=-34$. Therefore, if the given system has no solution, the value of $r$ is $-34$.