Answer
$t = 10.2~s$
Note that the smaller cube starts moving first.
Work Step by Step
We can find the minimum force required to make the smaller cube move:
$F = mg~\mu_s$
$F = (0.0020~kg)(9.8~m/s^2)~(0.65)$
$F = 0.01274~N$
Note that since the mass of the larger cube is greater, then a greater force is required to make it start moving.
We can find the required value of $q$ on each cube:
$\frac{kq^2}{r^2} = 0.01274~N$
$q^2 = \frac{(r^2)(0.01274~N)}{k}$
$q = \sqrt{\frac{(r^2)(0.01274~N)}{k}}$
$q = \sqrt{\frac{(0.060~m)^2(0.01274~N)}{9.0\times 10^9~N~m^2/C^2}}$
$q = 7.1386\times 10^{-8}~C$
We can find the required time for this charge to accumulate on each cube:
$t = \frac{7.1386\times 10^{-8}~C}{7.0\times 10^{-9}~C/s} = 10.2~s$
Note that the smaller cube starts moving first.
