jmc

Rewriting and reducing the fraction, we get $\frac{60}{2^3\cdot 5^8}=\frac{2^2\cdot 3\cdot 5}{2^3\cdot 5^8}=\frac{3}{2\cdot 5^7}$. Multiplying the numerator and denominator by $2^6$, we obtain $$\frac{3}{2\cdot 5^7}\cdot \frac{2^6}{2^6}=\frac{3\cdot 2^6}{2^7\cdot 5^7}=\frac{192}{10^7}=.0000192.$$Therefore, there are $\boxed{3}$ non-zero digits to the right of the decimal point.