As, $R=\rho l / A=\rho l /\left(\pi d^{2} / 4\right)$
Resistance for a given material is directly proportional to the length and inversely proportional to the square of diameter of round faces.
So for the second rod, diameter is $2.0 / 0.6=10 / 3$ times diameter of the first one. and, length is $1.0 \,mm / 1 \,m =1 / 1000$ times the length of first one.
Thus resistance of the second rod is,
$\frac{1 / 1000}{(10 / 3)^{2}}=9 \times 10^{-5}$ times the resistance of the first one.
So resistance of the second rod is
$3 \times 10^{-3} \times 9 \times 10^{-5}=2.70 \times 10^{-7} \,\Omega$