The answer does not lie on the surface (or does it?)

Find the mass out to a particular radius $a$ .

$dM = \rho \, dV = \rho \, 4 \pi r^2 \, dr = (3 - r) \, 4 \pi r^2 \, dr = 4 \pi \, (3 r^2 - r^3) \, dr \\ M(a) = 4 \pi \, \int_0^a (3 r^2 - r^3) \, dr = 4 \pi \Big ( a^3 - \frac{a^4}{4} \Big)$

Since the star is radially symmetric, it gravitates as though all of its mass is at the origin. Additionally, the mass farther out than $a$ makes no contribution to the gravitational field strength. The corresponding gravitational field strength is:

$g(a) = \frac{G M(a)}{a^2} = 4 \pi \, G \, \Big ( a - \frac{a^2}{4} \Big)$

To find the value of $a$ which extremizes this function, take the derivative and set equal to zero.

$1 - \frac{a}{2} = 0 \implies \boxed{a = 2}$

We can try a few other values of $a$ to confirm that this is indeed a maximum and not a minimum.

Addendum:

Here is some further justification of the statement "Since the star is radially symmetric, it gravitates as though all of its mass is at the origin. Additionally, the mass farther out than $a$ makes no contribution to the gravitational field strength".

Recall Gauss's Law for gravitation (here, the negative sign just means that the gravity pulls toward the origin - I ignored this earlier because it wasn't relevant):

$\int \int_s \vec{g} \cdot \, \vec{dS} = - 4 \pi G M$

Here, $M$ is the enclosed mass within the sphere of radius $a$ , naturally excluding the effects of mass not enclosed by the sphere. Since the mass distribution is radially symmetric, the gravity field is constant in magnitude over the sphere surface, and is everywhere parallel to the surface normal. Hence:

$\int \int_s \vec{g} \cdot \, \vec{dS} = \int \int_s g \, dS = g \, \int \int_s \, dS = g \, S = g \, 4 \pi a^2 = - 4 \pi G M(a) \\ g = -\frac{G M(a) }{a^2}$

It is therefore apparent that the solid sphere of mass gravitates as though it were a point-mass at the origin.

The answer didn’t specify a decimal point accuracy, so it must be an integer. It’s not 0, and must be less than or equal to 3. That leaves 3 options and 3 guesses. For me, it was my second guess that yielded the correct answer.

QED

Use same trick to solve this problem .

g=(4πG/r^2)Integral{ro(r')((r')^2)dr'},

substituting, ro(r')=3-r', and integrating

g=(4πG){r-(r/2)^2}, setting d(g)/dr=0,

Answer is r=2

Field at a distance $x$ (≤3) = $\int ^{x}_0 \frac {G(4πy^2dy)(3-y)}{x^2}$

= $k(x-\frac {x^2}{4})$ , k is some constant.

Maximum value of above function occurs at $x=2$

The mass inside a sphere with radius R  is given by $M=\int_0^R{4\pi r^2 \rho(r)}dr=\pi \int_0^R{12r^2-4r^3}dr= \pi (4R^3-R^4)$ . The gravitational field is $GM/R^2 =\pi G(4R-R^2)$ . Its maximum is where the derivative of this is 0, so when $4=2R$ , so at $\boxed{R=2}$ .