Answer
$Q=8.0\times 10^{-9}C$
Work Step by Step
According to given condition:
$mg=K\frac{Q^2}{d^2}$
As $d=2r$,
$mg=\frac{KQ^2}{(2r)^2}$.......eq(1)
We also know that:
$m=V\rho $
As the volume for sphere is $V=\frac{4}{3}\pi r^3$;
$\implies m=\frac{4}{3}\pi r^3 \rho$
Thus equation(1) becomes
$\frac{4}{3}\pi r^3 \rho g=\frac{KQ^2}{(2r)^2}$
This simplifies to
$Q=\sqrt{\frac{16\rho \pi g r^5}{3K}}$
We plug in the known values to obtain:
$Q=\sqrt{\frac{16\times 35\times 3.1416\times 9.8(1\times 10^{-2})^5}{3\times 8.99\times 10^9}}=8.0\times 10^{-9}C$