A positive cylindrical charge distribution of radius a and volume charge density rho is at the center of a thin grounded cylindrical conducting shell of radius b . What is the capacitance per unit length of this configuration?

The capacitance per unit length is C/L=Q/(VL) . So lets find the voltage of this configuration. In order to do that use Gausses' law to find E from the charge distribution. Make a Gaussian cylinder.
int int_A E*dr=Q_(enc)/epsilon_0
We know that we will enclose part of the charge distribution when 0ltrlta so make a Gaussian cylinder. Make a surface r between 0 and a.
E 2pi rL=(rho*pi r^2 L)/epsilon_0
E(r)=(rho r )/(2 epsilon_0)
0ltrlta.in the +r direction
Now we must find the field between altrltb . Make a cylindrical surface a distance r between a and b .
int int_A E*dr=Q_(enc)/epsilon_0
E (2pi r L)=(rho pi a^2 L)/epsilon_0
E(r)=(rho a^2)/(2epsilon_0 r)  in the +r direction for altrltb
Now since surface b is a grounded conductor it will take up an equal and opposite total charge. Therefore Gausses' law says that the E-field is zero for rgtb .
Since we now know E for all of space we can integrate for the potential of this system.
V=Delta V=V(b)-V(a)=-int_a^b E*dr=-int_0^a (rho r )/(2 epsilon_0)dr-int_a^b (rho a^2)/(2epsilon_0 r)dr
V=-(rho)/(4 epsilon_0 ) r^2|_0^a -(rho a^2)/(2 epsilon_0)ln(r)|_a^b
V=-[(rho)/(4 epsilon_0 )a^2+(rho a^2)/(2 epsilon_0)ln(b/a)]
V=[-(rho a^2)/(4epsilon_0)(2ln(b/a)+1)]
Although, V(b)ltV(a) here so for the purposes of this problem make V positive (can only have a positive C ).
V=Delta V=V(a)-V(b)=(rho a^2)/(4epsilon_0)(2ln(b/a)+1)
C/L=Q/(VL)=Q/((rho a^2)/(4epsilon_0)(2ln(b/a)+1)L)=Q/((Q a^2)/(pi a^2 L 4 epsilon_0)*(2ln(b/a)+1)L)
C/L=1/(4pi epsilon_0)(2ln(b/a)+1)^-1