This looks very much like a question about the orbital period of a planet (all your variables and the answer are correct for the planet Mars).
Which means your equation is slightly incorrect. Kepler's law of periods is:
<img src="http://www.codecogs.com/eq.latex?T^{2} = \frac{4\pi^{2}R^{3}}{GM}">
meaning that:
<img src="http://www.codecogs.com/eq.latex?T = \frac{2\pi\sqrt{R^{3}}}{\sqrt{GM}}">
Although, having said that, I might be misunderstanding where you finish your square root.
Either way, the answer is correct. I get:
<img src="http://www.codecogs.com/eq.latex?\begin{align*}T^{2}&=\frac{4\pi^{2}R^{3}}{GM}\\&=\frac{4.679\times10^{35}}{1.33\times10^{20}}\\&=3.52\times10^{15}\\T&=\sqrt{3.52\times10^{15}}\\T&=59313922.43\end{align}">
Which may or may not be what you were getting and although it looks way wrong it is in fact right. Kepler's law of period requires the variables to be in the following units:
M (mass of planet + mass of sun) has to be in Kg
R (orbital radius) has to be in m
G (gravitational constant) is given in units of m^(3) kg^(-1) sec^(-2)
Which when you put the units into the formula then the answer you get is actually in seconds.
So dividing 59313922.43 by 60x60x24 (i.e. converting it to days) gives the correct answer of 687 days.
If this question was from a maths textbook and it didn't describe the units of the variables you gave or mention that the formula for T will give an answer in seconds then it is a very harsh question. You obviously are not meant to know physics formulas for 2unit maths.
