Answer
$a = 1.9\times 10^{13}~m$
Work Step by Step
We can assume that the period of the comet is $~~1994-574~~$ which is $~~1420~years$
We can use Kepler's third law to find the semi-major axis $a$:
$a^3 = T^2$
$a = \sqrt[3] {T^2}$
$a = \sqrt[3] {(1420)^2}$
$a = 126.3~AU$
$a = (126.3~AU)(1.5\times 10^{11}~m/AU)$
$a = 1.9\times 10^{13}~m$
