How much does surface gravity of a star shift color and who observes the shift? Reading Richard Muller's new book Now he explains that since the gravitational field near the surface of star causes time to run slower relative to time further away from the surface, the frequency of the light is lower. So therefore the wavelength, color at the surface of the star would be redshifted.
What I'm confused about is this relative to an observer far away from the star? And as the observer approaches the star would the color shift towards red?
Is there a simple calculation to determine how much shift one would expect according to the star's mass?
 A: It's simpler than you (probably) think. Time runs more slowly near the star than it does far away. So if an observer near the star measures the period of some radiation to be $\tau$ then when that radiation reaches an observer far from the star that period will have some value $t$ that is greater than $\tau$. Since the period has increased the frequency has decreased i.e. the light has been red shifted.
For most stars (possibly excluding the densest most rapidly rotating neutron stars) the spacetime geometry around the star is well described by the Schwarzschild metric, and the time dilation is given by:
$$ \frac{\tau}{t} = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1} $$
In this equation $\tau$ is the time measured by an observer at distance $r$ from the centre of the star and $t$ is the corresponding time measured by an observer at infinity. In this case at infinity means far enough from the star for its gravity to be negligible. $G$ is the gravitational constant and $M$ is the mass of the star. (Strictly speaking the variable $r$ is only approximately the distance from the centre of the star, but we'll gloss over that).
Note that the right hand side is always less than one, so $\tau \lt t$ just as I described in my opening paragraph. Times measured near the star are shorter than the corresponding times measured far from the star.
And the change in the frequency is inversely proportional to the time dilation, so if some radiation is emitted at a distance $r$ with a frequency $\nu_r$ then the frequency at infinity is:
$$ \nu_\infty = \nu_r \sqrt{1 - \frac{2GM}{c^2r}} \tag{2} $$
It's all very well to talk about observers at infinity, but of course real observers are never at infinity so we might ask what happens for an observer at a large but finite value of $r$.
Suppose the light is emitted at $r = r_a$ and is being received by an observer at $r = r_b$, where $r_b \gt r_a$. Equation (1) gives us the time dilations for the two observers $a$ and $b$ relative to infinity:
$$ \frac{\tau_a}{t} = \sqrt{1 - \frac{2GM}{c^2r_a}} $$
and:
$$ \frac{\tau_b}{t} = \sqrt{1 - \frac{2GM}{c^2r_b}} $$
And to get the time dilation for $a$ relative to $b$ we just divide the two equations:
$$ \frac{\tau_a}{\tau_b} = \frac{\sqrt{1 - \frac{2GM}{c^2r_a}}}{\sqrt{1 - \frac{2GM}{c^2r_b}}} \tag{3} $$
And as before we end up with the red shift for radiation released at $a$ and received at $b$:
$$ \nu_b = \nu_a \frac{\sqrt{1 - \frac{2GM}{c^2r_a}}}{\sqrt{1 - \frac{2GM}{c^2r_b}}} \tag{4} $$
So if we start at $r = \infty$ and move inwards then the gravitational redshift, given initially by equation (2), decreases according to equation (4) and as we reach the source, i.e. $r = r_a$, the red shift disappears completely.
A: No, the other way around. Frequency gets redshifted as you go away from the star. And time goes faster. They go as the inverse of each other - think of it as the same number of cycles, if it is faster time, then it is fewer cycles per unit time. 
@John Rennie has the equations, you just need to interpret them right. From Eq (2) the freq at infinity is clearly less than at some distance r. Lower FRW at infinity means it got redshifted from the star to infinity. Or from the star to anywhere above it. So, a redshift going away from the star, and faster time. The opposite going into the star. 
As the observer approaches the star it gets bluer. As he goes away it gets redder. One simple equation, but I always have to think to decide which way it goes. 
A: 
Reading Richard Muller's new book Now he explains that since the gravitational field near the surface of star causes time to run slower relative to time further away from the surface, the frequency of the light is lower. So therefore the wavelength, color at the surface of the star would be redshifted.

There's a few issues with the above.
1) The degree of gravitational time dilation at some elevation relates to the gravitational potential at that location. 
2) The gravitational field at that location relates to the gradient in potential. 
3) The frequency of light doesn't actually change with altitude, in that E=hf and energy is conserved. Instead as Einstein said "an atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated". The light is emitted at a lower frequency. It doesn't reduce in frequency as it ascends. 
4) Colour appears redshifted when you're at the higher elevation, not the lower. This is because I do work on you when I lift you up. I add energy to you. Because of this it looks to you as if the ascending photon has lost energy. It hasn't. Instead you've gained it.   

What I'm confused about is this relative to an observer far away from the star? And as the observer approaches the star would the color shift towards red?

No, the observer would see an increasing blueshift as he approached. Or you might prefer to say the observed redshift diminishes. 

Is there a simple calculation to determine how much shift one would expect according to the star's mass?

Yes, use the gravitational time dilation expression $t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}}$ as per John Rennie's answer. 
