Answer
$2.2\times 10^{4} N$
Work Step by Step
According to $F=k\frac{Q_{1}Q_{2}}{r^{2}}$
The charges are $25µC = 2.5 \times 10^{-5} C$ and $2.5mC = 2.5 \times 10^{-3} C$, plug these two to $Q_{1}$ and $Q_{2}$.
The distance between them is $16cm = 0.16m$, plug it to r.
Then plug in k, the constant $9.0 \times 10^{9} N \times m^{2}/C^{2}$.
It's $F=9.0 \times 10^{9} N \times m^{2}/C^{2} \times \frac{ 2.5 \times 10^{-5} C \times 2.5 \times 10^{-3} C}{(0.16m)^{2}} = 2.2\times 10^{4} N$
