Is Your Charge Density Integration Correct for a Spherical Volume?

[SammyS]

First I did d(rho)/dr which is equal to 35.4*10^-12/R. Then I integrated d(rho) by which I got rho=35.4*10^-12. And then the last eqn will be q=rho V. But the answer was wrong.
I have a doubt on the formula I am using for E because that formula is for a point charge or a charged shell.

Back to your Original Post. (OP)
You are given the charge density, ρ, as a function of r.

If you differentiate that, with respect to r, and then integrate over r, you get charge density back - in some form. If you find an indefinite integral, and the evaluate the constant of integration, you simply get the charge density back as a function of r. If you evaluate the definite integral from 0 to R, as you did, you simply get the value of the charge density at r = R, i.e., at the surface of the sphere.

To find the charge over some volume you need to do a volume integral, as has been mentioned. In your case, with spherically symmetric charge distribution, the volume element that's handy to use is $dV=4\pi r^2dr$ .

Then $\displaystyle Q_{in}=35.4 (pC/m^3) \int_0^a \frac{r}{R} 4\pi r^2 dr$ is the amount of charge inside a sphere of radius $a$ for $a \le R$ .