Answer
$v = 121~m/s$
Work Step by Step
We can find the mass of the part of the sun that is inside Earth's orbital radius:
$M = \frac{\frac{4}{3}\pi(1.5\times 10^{11}~m)^3}{\frac{4}{3}\pi(5.90\times 10^{12}~m)^3}~(1.99\times 10^{30}~kg)$
$M = 3.27\times 10^{25}~kg$
We can find the Earth's orbital speed:
$\frac{M_E~v^2}{r} = \frac{G~M~M_E}{r^2}$
$v^2 = \frac{G~M}{r}$
$v = \sqrt{\frac{G~M}{r}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(3.27\times 10^{25}~kg)}{1.5\times 10^{11}~m}}$
$v = 121~m/s$
