Answer
(a) $9.454\times10^{15}~m$
(b) $6.30\times10^4$ AU in one light-year
Work Step by Step
(a) $1 ~year = (1 ~year)(365 ~\frac{days}{year})(24 ~\frac{h}{day})(3600 ~\frac{s}{h}) = 31536000 ~s$
$(2.998\times10^8 \frac{m}{s})(31536000 ~s) = 9.454\times10^{15}~m$
(b) $\frac{9.454\times10^{15}~m}{1.50\times10^{11}~m} = 6.30\times10^4$ AU in one light-year.
