The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)

is 18.0 mV at 25

use,
E = E0 - (0.059/n)*ln(Q)

here, E0 will be 0 because, both half cell is made up of Zn
and let M be the concentration of Zn+2 at cathode
Q = [Zn+2]anode/[Zn+2]cathode
= 0.100/M
n = 2

so,
E = 0 - (0.059/2)*ln(0.100/M)
18.0*10^-3 = - (0.059/2)*ln(0.100/M)
ln(0.100/M) = -0.6102
(0.100/M) = 0.543
M = 0.100/0.543
= 0.184
so,
required concentration = 0.184 M

Answer : 0.184 M