Find an approximated value for $\displaystyle \sqrt[ ]{9.03}$ using a Taylors polynomial of third degree and estimate the error.

I thought of solving it by using

$\displaystyle f(x)=\sqrt[]{x}$ centered at $\displaystyle x_0=9$

So

$\displaystyle P_n(x)=3+\dysplaystyle\frac{(x-9)}{6}-\dysplaystyle\frac{(x-9)^2}{216}+\dysplaystyle\frac{3(x-9)^3}{3888}$

Then I evaluated it at x=9.03, so I get:

$\displaystyle P_n(x)=3+\dysplaystyle\frac{(0.3)}{6}-\dysplaystyle\frac{(0.3)^2}{216}+\dysplaystyle\fra c{3(0.3)^3}{3888}\approx{3.049604167}$

I don't know if this is right, I've tried with the calculator and it gives 3.00500.... Now, how do I estimate the error? just by resting to the value the calculator gives the one I get?